Proof that the standard complex $d_{i+1} : E_{i+1} \to E_i$ is indeed a complex.

The problem statement is mostly quoting from Lang's Algebra, top of page 764. Let $S$ be a set. For $i = 0, 1, 2,\dots$ let $E_i$ be the free $\Bbb{Z}$-module generated by the $(i+1)$-tuples $(x_0, \dots, x_{i})$ with each $x_j \in S$. The tuples $x := (x_0, \dots, x_i)$ form a basis for $E_i$ over $\Bbb{Z}$. Define $d_{i+1} : E_{i+1} \to E_i$ on the tuples and the linear (or "homomorphic") extension to all of $E_i$ uniquely determines a $\Bbb{Z}$-module homomorphism: $$ d_{i+1}(x) = d_{i+1}(x_0, \dots, x_{i+1}) = \sum_{j = 0}^{i+1} (-1)^j (x^{\hat{j}} = (x_0, \dots, \widehat{x_j}, \dots x_{i+1})) $$ Our goal is to prove that $d^2 = 0$ or that $d_{i}\circ d_{i+1} = 0$ the zero map. We will use a trick of breaking up a summation into two summations over $k \lt j$ and $k \gt j$. In our notation, we can't have $k = j$ because $j$ was already removed from the original tuple $(x_0, \dots, \widehat{x_j}, \dots, x_{i+1})$. That is our notation doesn't re-index $x$ after removing the $j$th entry; I think this keeps things a lot simpler. $$ d_i \circ d_{i+1}(x) = d_i\left(\sum_{j = 0}^{i+1}(-1)^j (x^{\hat{j}}) \right) $$ But $d_i$ is linear on $\Bbb{Z}$-linear sums of tuples by definition, so we can distribute the $d_i$ into the sum: $$ = \sum_{j = 0}^{i+1}(-1)^j d_i(x^{\hat{j}}) $$ And now expanding next the $d_i$ and applying the above-mentioned trick: $$ = \sum_{j = 0}^{i+1}(-1)^j \left(\sum_{k=0 \\ k \lt j}^{i+1} (x^{\widehat{j,k}}) + \sum_{k=0 \\ k \gt j}^{i+1}(x^{\widehat{k,j}})\right) \tag{1} $$ Notice that our inner summation is up to $i+1$ and not $i$. We specifically do this (even though $d_i : E_i \to E_{i-1}$) because we're against re-indexing in order to complete this proof (see above note about that). Having said that, now we look at what happens to the sign of the summation terms when we have deleted the $j$th entry: When $k \lt j$ nothing should change about the sign of $(-1)^{k+j}$. But when $k \gt j$ we must add (or subtract) $1$ from the sign because of the inherent even/odd logic of consecutive numbers and we deleted an entry, giving the $k$th term an odd sign if it was originally even and vise versa. So the formula at (1) above is not correct. We've made a mistake! To correct it, write: $$ = \sum_{j = 0}^{i+1}(-1)^j \left(\sum_{k=0 \\ k \lt j}^{i+1} (x^{\widehat{j,k}}) - \sum_{k=0 \\ k \gt j}^{i+1}(x^{\widehat{k,j}})\right) $$ Notice the minus operator instead of the plus operator splitting the two inner sums. However, the sets of tuples (summation terms) $\{x^{\widehat{j,k}}: j,k =0, \dots, i+1\}$ and $\{x^{\widehat{k,j}} : k,j = 0, \dots, i+1\}$ are precisely the same thing. But because of that minus between the sums, the result is total cancellation or a value of $0$! $\blacksquare$