**Update:**
Now I know why my method fails. But I still wanna know how to work out the original question, that is to show the exactness of the chain complex $C_*(X)$ except for two positions $n=0,N-1$.

Here is a little problem about the normalization theorem in the theory of simplicial objects.

Given a finite set $X$ with $|X|=N>2$, we can construct a simplicial free abelian group $C_*(X)$ 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 $d(x_0,x_1,\dots,x_n)=\sum_{i=0}^n(-1)^i(x_0,\dots,\hat{x_i},\dots,x_n)$.
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 $n\neq 0,N-1$.

My strategy goes as follows: Consider the unnormalized complex $U_*$ with $U_n(X)$ defined to be the free abelian group generated by the unnormalized $(n+1)$-tuple $(x_0,x_1,\dots,x_n)$ of $n$ elements in $X$ which are allowed to have repetitions whose differentials are defined as above. Then an easy computation shows that the chain complex $U_*$ is acyclic for $n\geq1$(See for example Mac Lane's book *Homology*, Ex.1 in the end of Section 7, Chapter VIII, p. 238). And the normalization theorem(MacLane *Homology*, VIII.6, p.236 or the Theorem 3.3 in Normalization or thisQuestion) gives that the two chain complexes $C_*, U_*$ are quasi-isomorphic. So the acyclicity of $U_*$ implies the exactness of $C_*$. But $C_{N}=0, C_{N-1}$ is a free abelian group of rank $N!$ which injects into $C_{N-2}$ which also has rank $N!$. This forces the $(N-2)$-th differential $d_{N-2}:C_{N-2}\to C_{N-3}$ to be the zero map, which is impossible. I'm wondering which step goes wrong?