Geometric Proof of the Formula for Simplex Numbers

I was quite impressed by a beautiful proof¹ of the formula for n'th triangular number — it's kind of a bijective proof built on the top of geometric summation idea.

Question: does this argument generalize to higher dimensions? i.e. can someone prove the formula for n-th «k-dimensional simplex number» this way?

For example, the statement for k=3 is that n-th pyramidal number is $\binom{n+2}{3}$.

(One obvious fallback would be observing that any ball in the pyramid is characterized by its coordinates², i.e. balls are counted by a negative binomial coefficient — which is exactly the desired result. That's a good proof, but needless to say, that's not the proof I'm looking for.)

¹ original post by Mariano Suárez-Alvarez, cited by Vaughn Climenhaga

² k-simplex is the subset $x_0+\dots+x_k=1$, $x_i\ge0$ in $\mathbb{R}^{k+1}$, and n-th k-simplex number is the number of non-negative integer solutions of $x_0+\dots+x_k=n$

Place the balls so that the corners of the pyramid have coordinates (1,1,1), (n,1,1), (1,n,1), (1,1,n). Then for the ball at (x,y,z) we see that (x,x+y,x+y+z) is an ordered triple of distinct numbers between 1 and n+2 inclusive. There are $\binom{n+2}{3}$ such triples, and they are clearly in one-to-one correspondence with the balls. Geometrically, this corresponds to drawing planes x=constant, x+y=constant, and x+y+z=constant through a given ball, and seeing where these planes intersect the x axis.

(To compare with the case $k=2$, "straighten" the picture above into a right triangle. Then the diagonal lines become lines x=constant and x+y=constant, and the bottom row of balls (the blue ones) can be identified with the points where these lines intersect the x axis.)

This also works in higher dimensions, giving $\binom{n+k-1}{k}$.

Whether this proof really is any different from the one you suggested in the question can of course be debated...