Consider the space $X$ obtained as the quotient space of $\Delta^k$ when all faces of the same dimension are identified. Fix an ordering on the vertices of $\Delta^k$, say $v_0, \dots, v_k$. We compute the homology groups:

We have, for each $n$, $\Delta_n = \la e_n \ra \cong \mathbb{Z}$ where $e_n$ is the equivalence class containing all $n$-dimensional faces. WLOG, we can use $[v_0 \dots v_n]$ as a representative for $e_n$.

First, consider $n < k$. We have:

$$\delta_n(e_n) = \delta_n([v_0 \dots v_n]) = \sum_{i=0}^{n} (-1)^i [v_0 \dots \widehat{v_i} \dots v_n] = \sum_{i=0}^n (-1)^i e_{n-1} = \left\{ \begin{array}{lr} e_{n-1} & n \text{ is even} \\ 0 & n \text{ is odd} \end{array} \right.$$

And (note, $\delta_{n+1}$ is not the zero map, because $n < k$):

$$\delta_{n+1}(e_{n+1}) = \delta_{n+1}([v_0 \dots v_{n+1}]) = \sum_{i=0}^{n+1} (-1)^n [v_0 \dots \widehat{v_i} \dots v_{n+1}] = \sum_{i=0}^{n+1} (-1)^n e_n = \left\{ \begin{array}{lr} 0 & n \text{ is even} \\ e_n & n \text{ is odd} \end{array} \right.$$

So, when $n$ is even, we have $\ker \delta_n = \set{0}$ and $\im \delta_{n+1} = \set{0}$. When $n$ is odd, we have $\ker \delta_n = \la e_n \ra$, and $\im \delta_{n+1} = \la e_n \ra$. Then, in both cases, $H_n(X) \cong \set{0}$.

Now consider $n=k$. If $k$ is even, then $\ker \delta_k = \set{0}$, by the same argument as above. Since $\delta_{k+1}$ is the zero map, $\im \delta_{k+1} = \set{0}$ and we have $H_k(X) = \set{0} / \set{0} \cong \set{0}$. However, if $k$ is odd, we have $\ker \delta_n = \la e_k \ra$, by the same argument as above. But, since $\delta_{k+1}$ is still the zero map, we have $\im \delta_{k+1} = \set{0}$ and $H_n(X) = \la e_n \ra / \set{0} \cong \mathbb{Z}$. To summarize,

$$H_n(X) = \set{0}, n < k \qquad H_k(X) = \left\{ \begin{array}{lr} \set{0} & k \text{ is even} \\ \mathbb{Z} & k \text{ is odd} \end{array} \right.$$