The singular boundary squares to zero

If $n=0$, then ${\mathrm{\partial }}_0=0$ by [L1], so the degree-zero boundary is already zero. If $n=1$ and $\sigma :{\mathrm{\Delta }}^1\to X$ is a singular $1$-simplex, then ${\mathrm{\partial }}_0{\mathrm{\partial }}_1\sigma ={\mathrm{\partial }}_0(\sigma {\delta }_0-\sigma {\delta }_1)=0,$ again because ${\mathrm{\partial }}_0=0$. Thus the claim holds in the two low degrees.

Assume $n\ge 2$ and let $\sigma :{\mathrm{\Delta }}^n\to X$ be a singular $n$-simplex. Expanding twice with [L1] gives ${\mathrm{\partial }}_{n-1}{\mathrm{\partial }}_n\sigma ={\sum }_{i=0}^n{\sum }_{j=0}^{n-1}(-1{)}^{i+j}\sigma {\delta }_i{\delta }_j\mathrm{.}$ Reindex the terms by pairs $0\le j<i\le n$ and $0\le i\le j\le n-1$. By [L2], the term $\sigma {\delta }_i{\delta }_j$ with $j<i$ equals $\sigma {\delta }_j{\delta }_{i-1}$, but the two appearances carry opposite signs because $(-1{)}^{i+j}=-(-1{)}^{j+(i-1)}\mathrm{.}$ Hence every codimension-two face occurs twice with opposite coefficients, so the whole sum is zero.

Since singular simplices generate $C_n(X;\mathbb{Z})$ and the coefficient version is obtained by tensor extension, steps 1.1 and 1.2 imply ${\mathrm{\partial }}_{n-1}{\mathrm{\partial }}_n=0$ on $C_n(X;G)$ for every $n\ge 1$, with degree $0$ handled separately in step 1.1.