Let $G$ be a group and, for $n \geq 1$, let $B_{n}$ be the free $G$-module with basis $X=\left\{\left[x_{0}|\cdots| x_{n}\right]: x_{i} \in G\right\}\left(B_{n}\right.$ is the $n$th term of the bar resolution). Prove that $\left(B_{n}\right)_{G}=B_{n} / \mathcal{G} B_{n}$ is the free $\mathbb{Z}$-module with basis $X$.