The boundary operator squares to zero

It suffices to check the claim on a single $n$-simplex $\sigma = [v_0, \dots, v_n]$ and extend by linearity. By definition, $\partial_n \sigma = \sum_{i=0}^n (-1)^i [v_0, \dots, \hat{v_i}, \dots, v_n]$, where $\hat{v_i}$ means $v_i$ is deleted.

Applying $\partial_{n-1}$ to each term and deleting a second vertex $v_j$ (with $j \ne i$) from $[v_0, \dots, \hat{v_i}, \dots, v_n]$ produces the face $[v_0, \dots, \hat{v_j}, \dots, \hat{v_i}, \dots, v_n]$ that is missing exactly the two vertices $v_i, v_j$. Carefully tracking the sign: when $j < i$ the face is deleted at position $j$ first (sign $(-1)^j$) inside a term already carrying sign $(-1)^i$, giving total sign $(-1)^{i+j}$; when $j > i$, deleting $v_j$ from the $(n-1)$-simplex $[v_0,\dots,\hat{v_i},\dots,v_n]$ removes the vertex now sitting at position $j-1$ (because $v_i$ was already removed), giving sign $(-1)^i (-1)^{j-1} = -(-1)^{i+j}$.

So $\partial_{n-1}\partial_n \sigma = \sum_{j<i} (-1)^{i+j} [\dots,\hat{v_j},\dots,\hat{v_i},\dots] + \sum_{j>i} \left(-(-1)^{i+j}\right) [\dots,\hat{v_i},\dots,\hat{v_j},\dots]$. Every face missing exactly two vertices $\{v_i, v_j\}$ with $i < j$ appears exactly twice in this double sum: once from deleting $v_i$ then $v_j$ (contributing $-(-1)^{i+j}$) and once from deleting $v_j$ then $v_i$ (contributing $+(-1)^{i+j}$), and these two contributions are exact opposites.

Every term cancels in pairs, so $\partial_{n-1}\partial_n \sigma = 0$ for every simplex $\sigma$, hence $\partial_n \circ \partial_{n+1} = 0$ on all of $C_n$ by linearity.