(elementary?) proof of simplicial homology groups of $\Delta_n$ Using the (trivial) CW-complex structure of $\Delta_n$, I would like to compute the homology groups of $\Delta_n$. It's obvious that (for $k \leq n$) $C_k \simeq \mathbb{Z}^{{n+1 \choose k+1}}$, and that $im \delta_{k+1} \subset ker \delta_k$. It's the reverse inclusion for $k \geq 1$ that I'm not able to show.
I know how to prove this using homotopy invariance of homology, as $\Delta_n$ is contractible, and a proof in that vein is given in a comment here Explicit calculation of simplicial homology - but I'm wondering if there is something more elementary? The CW-complex structure makes computing the homology groups of the sphere almost trivial, and I was hoping that it would be easy for the disk as well. 

It's quite easy in simplicial homology to write down a contracting homotopy
for the identity map of $C(\Delta_n)$. Let the vertices 
of $\Delta_n$ be labelled $0,1,\ldots,n$.
Then the $k$-chains are labelled by tuples $(j_0,\ldots,j_k)$ where
$0\le j_0<j_1<\cdots< j_k\le n$. Define $T:C_k(\Delta_n)\to C_{k+1}(\Delta_n)$
by $T:(j_0,\ldots,j_k)\mapsto(0,j_0,\ldots,j_k)$ if $j_0\ge1$
and $T:(j_0,\ldots,j_k)\mapsto0$ if $j_0=0$.
If then $\delta_k(\gamma)=0$ for $\gamma\in C_k$ (with $k\ge1$) then
$\gamma=\pm\delta_{k+1}(T(\gamma))$ (I can't be bothered to check the sign),
so $\ker\delta_k\subseteq\mathrm{im}\,\delta_{k+1}$.
