I'm learning about group cohomology from Knapp's book Advanced Algebra. Given a group $G$, and an abelian group $M$, he defines $F_n$ to be the $(n+1)$-fold product of $G$, endowed with a $G$-action by $g\cdot(g_0, g_1, \dotsc, g_n)= (gg_0, gg_1, \dotsc, gg_n)$. Then he defines boundary maps $F_n \to F_{n-1}$ and applies $\hom_{\mathbb{Z}G}(-, M)$ to get a cochain complex. He points out that $F_n$ is free as a $G$-module over the $(n+1)$-tuples with $1$ in the first entry. Further, $\hom_{\mathbb{Z}G}(F_n, M)$ is isomorphic as an abelian group to $C^n(G, M)$, i.e. the abelian group of set maps from $G^n$ to $M$, via the isomorphism $$\hom_{\mathbb{Z}G}(F_n, M) \overset{\Phi}{\to} C^n(G,M)$$ $$\Phi(\varphi)(g_1, \dotsc, g_n) = \varphi(1, g_1, g_1g_2, \dotsc, g_1\dotsb g_n).$$
My Question: Why does he choose this isomorphism? I'm not seeing the point of the successive multiplication in later entries. It seems like $$\Phi(\varphi)(g_1, \dotsc, g_n) = \varphi(1, g_1, \dotsc, g_n)$$ would work. What am I missing?