Proof that the standard complex is indeed an exact sequence.

See this blog post which proves that $d^2 = 0$ or indeed $\operatorname{im} d_{i+1} \subset \ker d_i$. Our goal here is to prove the opposite inclusion. By a trick mentioned in Lang, we can freely choose any element $z \in S$ and define $h: E_{i} \to E_{i+1}$ by linearly extending $h(x_0, \dots, x_i) = (z, x_0, \dots, x_i)$. We then need to prove that $dh + hd = \text{id}$ which would then imply that if $x \in \ker d$, then $x = (dh + hd)(x) = dh(x)$ or that $x$ is in the image of $d$. First we must get the chain map indices correct instead of succinctly working with the plain "$d$" notation: is what we think is meant by "$dh + hd$". So we need to prove that $d_{i+1} h_i + h_{i-1} d_i = \text{id}$. So here we go: $$ (d_{i+1} h_i + h_{i-1} d_i)(x) = d_{i+1} h_i(x) + h_{i-1} d_i(x) = \\ \sum_{j=0}^{i+1}(-1)^j(z, x_0, \dots,\widehat{x_j},\dots, x_{i}) + h_{i-1}\left(\sum_{j=0}^{i}(-1)^j(x_0, \dots, \widehat{x_j}, \dots, x_{i})\right) = $$ Notice that the first summation has $i+2$ terms while the second summatio has $i+1$ terms. The odd man out is when $j = 0$ or $\widehat{z}$ is removed as an entry. That leaves $(x_0, \dots, x_i)$. The rest of the terms get cancelled by a term of opposite sign in the second summation since everything is shifted by $1$! $\blacksquare$