commutative law of monoid

Question

can someone help me understand proof of this property? Let G be a commutative Monoid and $x_1,..,x_n \in G$. Let $\psi$ be a bijection of the set of intergers $\{1,2,3,...,n\}$ . Then $$ \prod^n_{v=1}x_{\psi(v)}=\prod^n_{v=1}x_v$$ Assume for $n-1$, let $k$ be the integer s.t $\psi(k)=n$, $$ \prod^n_{v=1}x_{\psi(v)}=\prod^{k-1}_{v=1}x_{\psi(v)}*x_{\psi(k)}\prod^{n-k}_{v=1}x_{\psi(v+k)}$$ Define a map $\phi(v) = \psi(v)$ if $ v<k $, else $\phi(v)=\psi(v+1)$ on $\{1,2,...,n-1\}$ Then, $$ \prod^n_{v=1}x_{\psi(v)}=\prod^{k-1}_{v=1}x_{\phi(v)}*x_{\phi(k)}\prod^{n-k}_{v=1}x_{\phi(v-1+k)}\\=\prod^{n-1}_{v=1}x_{\phi(v)}x_n$$ I'm confused on why do we need to define the map $\phi$ in this proof? I feel like the definition of $\phi$ is redundant. since k-1, n-k < n, and shouldn't the property hold by applying the hypothesis to $\psi$? Thanks!