[bojan]

Guys, its like this, without any mathematics (and assuming we are dealing with ideal optics - no absorption of light in glass or on reflective surfaces):

Firstly, the light from the objects is collected by the aperture of the objective.
So, in general, wider the aperture, brighter the image formed by objective.

Secondly, from this collected light the image is formed at the focal plane of the objective.
Further away the focal plane is (longer the focal length), the larger the image. Now, in case of nebulae (spread or extended objects) - all the available amount of light, collected by aperture is spread over the image area and if the image area is wider (longer focal length), the surface brightness of the nebula image (photons per square mm or whatever) will be lower (same amount of light on the wider image).
For very small objects (including stars, which are almost point sources) if the size of their image is smaller than the size of the sensor (pixel, halogen particle, rod in the eye), the surface brightness will not change because all the available light is concentrated on that one sensor or point.
If we go into mathematical analysis of the this situation, it can be shown that the surface brightness of the formed image of the spread objects depends only on the ratio between focal length and aperture diameter (assuming aperture is round, and yes, it is because of the square low). However, their total, integrated brightness depends on the aperture (area of the objective.. because it is the aperture that collects the light).

For visual telescope, the logic is the same, with the difference we have the magnification as a factor (higher magnification, wider spread - lower surface brightness of the extended objects, but point-like sources (stars) will not be affected, until the diffraction effects take place)
Ant that's all there is.. easy