Yes; one way to see this is with a little calculus. It all boils down to the fact that the indefinite integral of x^(n-1) is x^n / n. (That's where 1/n comes in.)
Suppose we have an n-dimensional pyramid. It has an (n-1)-dim'l base with volume B, and that base tapers to a point; for simplicity, it has height 1.
Take cross-sections as we travel from the base to the height. How big is a cross section at height y?
Each cross section is a (n-1)-dim'l shape, and its volume is B * y^(n-1) because this is how volumes scale in dimension n-1. [Examples: In 2D if you x2 the sides of a shape, it's "volume" (area) is x4. In 3D if you x2 the sides of a shape, it's volume is x8; in 4D it would be x16, etc.]
Now take the integral to add all of these cross-sections:
integral(from 0 to 1 of B * y^(n-1))
= B * y^n / n evaluated from 0 to 1
= B / n.
If the shape is scaled along the height dimension by h, then we get the more general formula:
Suppose we have an n-dimensional pyramid. It has an (n-1)-dim'l base with volume B, and that base tapers to a point; for simplicity, it has height 1.
Take cross-sections as we travel from the base to the height. How big is a cross section at height y?
Each cross section is a (n-1)-dim'l shape, and its volume is B * y^(n-1) because this is how volumes scale in dimension n-1. [Examples: In 2D if you x2 the sides of a shape, it's "volume" (area) is x4. In 3D if you x2 the sides of a shape, it's volume is x8; in 4D it would be x16, etc.]
Now take the integral to add all of these cross-sections:
integral(from 0 to 1 of B * y^(n-1)) = B * y^n / n evaluated from 0 to 1 = B / n.
If the shape is scaled along the height dimension by h, then we get the more general formula:
volume = h * B / n.