TITLE: Simplicial homology of the skeleton of a simplex
QUESTION [8 upvotes]: Let $n$ and $k$ two natural numbers. We consider the (abstract) simplicial complex $K$ on  $n$ vertices $v_1,\dots,v_n$ and such that a subset of $\{v_1,\dots,v_n\}$ is a face of $K$ if and only if it has at most $k$ elements. So $K$ is the $(k-1)$-skeleton of a $(n-1)$-simplex. 
What is the homology of this complex? On various examples, I can see that every homology group is $0$ except the last one (and the zeroth group). I guess it is a well-known result but I have no idea on how to prove it. Direct computations in the general case are rather tedious.

REPLY [6 votes]: It's just a matter of arithmetic, I think. Observe that the $n$-simplex has $\binom{n+1}{r+1}$ $r$-simplices as faces. Thus, the Euler characteristic of the simplicial chain complex of the $k$-skeleton is
$$\sum_{r = 0}^{k} (-1)^r \binom{n+1}{r+1} = 1 + (-1)^k \binom{n}{k+1}$$
and as you say, its homology is concentrated in degrees $0$ and $k$, so (assuming $k \ne 0$), we have
$$b_0 + (-1)^k b_k =  1 + (-1)^k \binom{n}{k+1}$$
where $b_r$ is the $r$-th Betti number, hence:
$$b_k = \binom{n}{k+1}$$