A Classic Paradox: Frankly, this topic may better be omitted for beginners, and instruction sources always do skip it. Why this topic is true really does not affect usage this time. It is sort of a distracting complication, and it should be enough to just know that camera distance does not affect exposure. But we may still be puzzled about why it doesn't? Certainly flash to subject distance does fundamentally affect flash exposure according to the inverse square law. When examining results, camera distance obviously seems NOT affected by the inverse square law (nevertheless, any and all distance is affected, which makes this case be very puzzling).
Camera distance obviously can change the scene seen in the frame, which affects metering and camera settings, which is not what this topic is about. This topic says the same settings will reproduce the mountain at the same brightness level, regardless if it is at the horizon, or we are standing on it.
This subject is a diversion from the fundamentals, and it is isolated here to make it easy to skip over it. This is an advanced topic, not essential, perhaps complicated, requires some thought, philosophical even, and it is probably just a distraction better left out here - because many people are not interested in why, but other people have argued this for decades. (Don't send arguments to me, I've been there, done that). But people do have this question, and it is a great question. Just not easy to answer. I am not so much encouraging pondering this subject, while the answer is important, the reason is nothing we must know. I'm merely offering this for those that wonder.
First, any theory with a conclusion that exposure does depend on camera distance is obviously simply wrong. You see what happens. No matter how far from the mountain we stand with our camera, the answer is always Sunny 16. We know the answer, we only wonder why.
To start with, the mountain is not a light source. It merely reflects the light from the Sun (the source). No matter where on earth we stand, the Sun is always 93 million miles away, or 150 million kilometers.
Light appears dimmer at greater distances from the light source. The light from a flashlight, or our car's headlights, for example. Stars appear to be an infinitesimal fraction of our sun's brightness - Stars appear relatively dim only because of their astronomical distance. This is the Inverse square law.
Yet a photon (elementary particle which carries this light energy) remains a photon, full strength, unaffected no matter how many millions of light years it has traveled. It is always a photon.
What does change as the light's coverage angle spreads out with distance is the relative density of photons, per unit of area in the image we view. Intensity is the light per unit of area. As the distance increases, the angle and the same photons spread out to cover a larger area, photons now farther apart, diluted (by distance), so to speak. We "see" less energy per unit of area, and this then appears dim.
But the subject as seen by the camera at greater distance, the lighted object area also appears smaller. When ten times more distant, the subject dimensions are only 1/10 size, which is 1/10 x 1/10 = 1/100 the area. Inverse square law says the light is 1/100 as bright at 10 times distance. So 1/100 the light in 1/100 the area is the same apparent intensity, per unit of area. It exactly balances out.
The camera sees the same exposure (of this subject area), regardless of distance. It is because of the Inverse Square Law that this is true, there is no conflict.
OK, but while standing there, what if then you zoom in? That image becomes larger, why isn't it brighter?
Right, a focal length 2x longer (100 mm vs 50 mm) does make the subject appear 2x larger, except our zoom crops our view into 1/4 the area still visible. We see 1/4 the light in 1/4 the area, which is the same light per unit of area.
But maybe we also shift our view to look at something twice as far (with the 2x focal length). Why isn't that more dim? The same exposure is because of the light per unit area stated before.
But the f/stop is explained this way: f/stop is defined as focal length / aperture diameter, so 2x more focal length is a 2x wider aperture diameter for the same f/stop number. This larger aperture diameter is 4x more window area to let in light. This balances out exactly again, 4x more light window area to balance the 1/4x smaller "light per unit area" that our lens sees. This is why f/stop is defined that way, so f/4 is the same f/4 exposure on any lens of any size.
That's it, but said again with more words:
There are two cases: 1) light that is incident onto the subject, from the light source, illuminating the subject.
And there is 2) light reflected from the subject, entering the camera lens (the subject itself is not a light source).
Distance from the incident light source to the subject definitely affects exposure of that subject, due to the inverse square law. Simply because, at a greater distance, the angular size of the subject obviously appears a smaller target, and less of the light's total output hits it. We know that if subject is 10x more distant from the light source, then it appears smaller, and its apparent size is 1/10 of the width and 1/10 of the height (similar triangles), or 1/100 of the apparent area when closer. This 1% area intercepts only 1/100 as much light hitting it. Seeing only 1/100 the light due to 10x the distance is the definition of the inverse square law.
Subject distance from the camera does not affect exposure (of the light reflected from that subject), which might seem as if the inverse square law does not apply to this reflected light. However, it does, exceptions are simply not possible - but it is seen in a different way. As previously explained, inverse square law simply applies to the geometry of angles spreading in space, and is about the distance and spread in space, and is not about any actual property of light at all. Camera distance is no exception, but we see it as a special case, and it is pretty tricky. Camera distance and zoom may affect the overall view the camera sees, but it does not affect the apparent brightness of any smaller part of that view.
For example, when photographing something from 10 feet (say an actor on a lighted stage), and again from 100 feet (all else the same), the subject's reflected light is the same. We can see this with our eyes, and it photographs that way too. But the size of the subject is not the same in the camera's view. A subject is an area, and its image covers an area on the camera sensor, or it covers an area on the cones in our eye. The greater its distance, the smaller it appears, and the less area it covers at our end. But we use the same exposure, even if more distant, so the path to camera might seem to be unaffected by the inverse square law - except ironically, it is the inverse square law that makes this so.
For why, suppose at close range, the subject fills the camera sensor. But at 10x distance, with the same setup, the angular size of the image in our lens is a tiny target now. At 10x distance, the object size appears tiny, and now only fills 1% of the sensor (1/10 of the height, 1/10 of the width). Where once (for example) 100x100 pixels (=10,000) was some feature of detail, now the same feature only covers 10x10 pixels (=100, 1%). The scene might appear more dim with distance, because all of its light is concentrated into a smaller area (which smaller subject is the same illumination). Only 1% of 10,000 = 100 pixels are the same brightness as before. So 1/100 less light, but concentrated in corresponding smaller area of 1/100 area, is the same brightness level and exposure, per unit of area. The inverse square law is what makes this be true, the reason why camera distance is not a factor of exposure.
Light reflected from the subject is a different situation than light incident on the subject. So the inverse square law is seen in a different way. Camera distance does not change the illumination on the subject, but we view it in a different way.
Suppose we used a 50 mm lens at f/4 before. Now we switch to a 500 mm lens at f/4, still at 10x distance. But 10x focal length magnifies, and means we will continue to see this subject as full frame size again, even at 10x distance. The aperture diameter at f/4 is no longer 50 mm/f4, but now is 500 mm/f4, which is 10 times larger aperture diameter, and 100 times larger aperture area, letting in 100 times more light, and so 10x greater distance appears the same brightness again - at the same f/4.
The difference in these paradoxical views is that camera distance automatically compensates and balances these two factors (image size and inverse square law intensity), but flash distance only affects illumination of the subject, and does not affect the camera view or image size. There have been endless philosophical arguments about this paradox - people discovering it do like to debate why.
But the actual result is undebatable, and the bottom line remains: We clearly see that the distance to the camera is not a factor of exposure. For extreme example, the Moon is illuminated by the same Sun we use, at basically the same distance. So a photograph of the Moon from Earth requires the same Sunny 16 exposure as when the astronauts were standing on the Moon (note that the overall albedo of the Moon is about the same as a 12% gray card. It is darkish, and only appears bright in the black sky). Note though, full moon is flat frontal lighted, half moon is side lighted, and new moon is rear lighted, so this lighting factor does cause a little change.
However, distance to the light source definitely is a major factor of exposure, extremely critical to realize for flash. The inverse square law tells us how much effect with distance, we can predict and plan it. Every flash shot is concerned with (a) trying to arrange all of the subject to be at the same distance from the flash, and (b) in some degree, controlling the two exposures of both flash and continuous ambient light with shutter speed (Part 4 here). This realization is the difference between knowing and no clue.