I am following the discussion of the Koszul complex in Lang's "Algebra".
Suppose $ R $ is a commutative ring, $ S $ is an $ R-$module, and $ f_{1},\dots,f_{r} \in R. $
Define:
$ K_{0}(f) = R. $
$ K_{1}(f) $ is a free module with basis $ \lbrace g_{1} \wedge \dots \wedge g_{r} \rbrace. $
$ K_{p}(f) $ is a free $ R-$module $ \bigwedge^{p}S $ with basis $ \lbrace g_{i_1} \wedge \dots \wedge g_{i_p} \rbrace $ where $ 1 \leq i_{1} < \dots < i_{p} \leq r. $
$ K_{r}(f) $ is a free $ R-$module $ \bigwedge^{r}S $ of rank $ 1 $ with basis $ g_{1} \wedge \dots \wedge g_{r}. $
Let $ d_{p} $ be the boundary map between $ K_{p}(f) $ and $ K_{p-1}(f). $ That is $$ d_{p} : K_{p}(f) \rightarrow K_{p-1}(f), $$ and
$$ d_{p}(g_{i_1} \wedge \dots \wedge g_{i_p}) = \sum_{j=1}^{p} (-1)^{j-1}x_{i_j}g_{i_1} \wedge \dots \wedge \widehat{g_{i_j}} \wedge \dots \wedge g_{i_{p}}. $$
I am having difficulty showing that $$ d_{p-1}\Bigg( \sum_{j=1}^{p} (-1)^{j-1} x_{i_j}g_{i_1} \wedge \dots \wedge \widehat{g_{i_j}} \wedge \dots \wedge g_{i_{p}} \Bigg) = 0. $$
To see that $d\circ d=0$, think about it this way. You can remove two elements $i_j, i_k$ from the set $\{i_1,\ldots,i_p\}$ by removing first $i_j$ and then $i_k$ or vice versa. You will get the same coefficient $x_{i_j}x_{i_k}$, but the sign will be different. Therefore the two terms arising in this way cancel.