Given a finite set $X$ with $|X|=N>2$, we can construct a simplicial free abelian group $C_*(X)$ (which is a chain complex) defined as follows: for each $n\geq0$, $C_n$ is defined to be the free abelian group generated by the set of all $(n+1)$-tuples $(x_0,x_1,\dots,x_n)$ of distinct elements in $X$; and set $C_n(X)=0$ for $n\geq N$. The differential map is defined as $d(x_0,x_1,\dots,x_n)=\sum_{i=0}^n(-1)^i(x_0,\dots,\hat{x_i},\dots,x_n)$ where the hat on $x_i$ means it's omitted.
An exercise in Weibel's book (see Weibel The K-Book: An introduction to algebraic K-theory, VI.5, Ex.5.1., p.550 or p.35) asserts that $H_n(C_*)=0$ for $n\neq 0,N-1$.
I'm wondering how to show it?
As indicated in Weibel's book, this argument is due to K. Hutchinson.
For $x \in X$, define the operator $S_x: C_n(X) \to C_{n+1}(X)$ defined on generators by $$S_x(t_0, t_1, \ldots, t_n) = \begin{cases} (x, t_0, t_1, \ldots, t_n), & \text{if } x \notin \{t_0, t_1, \ldots, t_n\} \\ 0, & \text{otherwise}. \end{cases}$$
We calculate: $$dS_x(t_0, t_1, \ldots, t_n) = \begin{cases} (t_0, t_1, \ldots, t_n) - S_x d(t_0, t_1, \ldots, t_n), & \text{if } x \notin \{t_0, t_1, \ldots, t_n\} \\ 0, & \text{otherwise}. \end{cases}$$ In either case, $(1 - dS_x)(t_0, t_1, \ldots, t_n)$ is contained in the subgroup spanned by generators that contain the element $x$.
Given a subset $I \subseteq X$, write $D(I) \subseteq C_\bullet(X)$ for the subgroup spanned by generators that contain every element of $I$. The conclusion of the previous paragraph is that $(1 - dS_x)z \in D(\{x\})$ for any $z \in C_n(X)$. In fact, if $z \in D(I)$ and is a cycle, then $(1 - dS_x)z \in D(I \cup \{x\})$.
Now let $z$ be an $n$-cycle. We have $\left(\prod_{x \in X} (1 - dS_x) \right)z \in D(X)$. However, if $n + 1 < \lvert X \rvert$, then $D(X) \cap C_n(X) = \{0\}$. So $$\left(\prod_{x \in X} (1 - dS_x) \right)z = 0.$$ This can be rewritten as $z = dw$ for some $w$, so $z$ is a boundary. Hence the chain complex $C_\bullet(X)$ has trivial homology is degrees strictly between $0$ and $\lvert X \rvert - 1$.