Let me just put my comments as an answer: (i) To compute the cohomology groups $H^n(G,A)$, for $A$ a (left) $G$-module and where $$H^n(G,A)=\text{Ext}^n_{{\mathbb Z}G}({\mathbb Z},A)$$ one needs a projective resolution of the $G$-module ${\mathbb Z}$, say $$\cdots\stackrel{d_3}\rightarrow P_2\stackrel{d_2}\rightarrow P_1\stackrel{d_0}\rightarrow P_0\stackrel{\varepsilon}\rightarrow {\mathbb Z}\rightarrow 0$$ such that $$H^n(G,A)=\text{Ext}^n_{{\mathbb Z}G}( {\mathbb Z},A)=\ker(d^*_{n+1})/\text{Im}(d_n^*).$$
(ii) One choice of projective resolution of ${\mathbb Z}$ is given by the free ${\mathbb Z}$-modules $P_n$ with basis the $(n+1)$-tuples $(g_0,\ldots,g_n)$ of elements of the group $G$ and where the action of $G$ is by (left) translation: $g(g_0,\ldots,g_n)=(gg_0,\ldots,gg_n)$. It follows that the $P_n$ are free $G$-modules with basis the $(n+1)$-tuples of the form $(1,g_1,\ldots,g_n)$. The $G$-morphisms $$d_n:P_n\rightarrow P_{n-1}$$ are defined on the generators by the formula $$d_n(g_0,\ldots,g_n)=\sum_{j=0}^n(-1)^n(g_0,\ldots,\widehat{g_j},\ldots,g_n)$$ where $\widehat{g_j}$ means that this term is omitted. The morphism $\varepsilon$ sends a generator $(g_0)$ to $1$. One shows that the $(P_n,d_n)$ form indeed an exact sequence and upon appliying the functor $\text{Hom}_{G}(-,A)$ one obtains a complex whose cohomology groups are $H^n(G,A)$. All of these is pretty standard fare in any homological algebra textbook. The important point is now to observe that an $n$-cochain $f\in \text{Hom}_G(P_n,A)$ is determined by its values in the generators $(g_0,\ldots,g_n)$ of $P_n$ and since the cochain is a $G$-module morphism (we say that it is $G$-covariant), then $$f(g_0,g_1,\ldots,g_n)=f(g_0(g_0^{-1}g_1,\ldots,g_0^{-1}g_n))=g_0f(1,g_0^{-1}g_1,\ldots,g_0^{-1}g_n)$$ i.e., $f$ is determined by its values on the generators of $P_n$ of the form $(1,g_1,\ldots,g_n)$ and then one may also consider another projective resolution of ${\mathbb Z}$ given by the free $G$-modules $Q_n$ with generators $[g_1,\ldots,g_n]$ (and where $Q_0$ is the free $G$-module generated by a symbol $[\;]$. One then shows that $P_n\simeq Q_n$ as $G$-modules and defining $$\delta_n:Q_n\rightarrow Q_{n-1}$$ by $$\delta_n[g_1,\ldots,g_n]=g_1[g_2,\ldots,g_n]+\sum_{j=1}^{n-1}(-1)^j[g_1,\ldots,g_jg_{j+1},\ldots,g_n]+(-1)^n[g_1,\ldots,g_{n-1}]$$ the $(Q_n,\delta_n)$ form a free $G$-resolution of ${\mathbb Z}$ and there is an isomorphism of chain complexes $\varphi:(P_n,d_n)\rightarrow (Q_n,\delta_n)$, answering your question. Let just add some details: First, the morphisms $\varphi_n:P_n\rightarrow Q_n$ are given by $\varphi_n(g_0,\ldots,g_n)=g_0[g_0^{-1}g_1,g_1^{-1}g_2,\ldots, g_{n-1}^{-1}g_n]$ and the corresponding $\psi_n:Q_n\rightarrow P_n$ are $\psi_n[g_1,\ldots,g_n]=(1,g_1,g_1g_2,g_1g_2g_3,\ldots, g_1g_2\cdots g_n)$. A lenghty, but straightforward, computation shows that $\delta_n=\varphi_{n-1}\circ d_n\circ \psi_n$.