Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either

Splitting law, base $n=0$: the left side is ${\prod }_{i<m+0}{g}_i={\prod }_{i<m}{g}_i$, and the right side is $\left({\prod }_{i<m}{g}_i\right)\cdot {\prod }_{j<0}{g}_{m+j}=\left({\prod }_{i<m}{g}_i\right)\cdot e={\prod }_{i<m}{g}_i$; so the two agree, for every $m$.

Splitting law, inductive hypothesis: fix $n$ and assume ${\prod }_{i<m+n}{g}_i=\left({\prod }_{i<m}{g}_i\right)\cdot \left({\prod }_{j<n}{g}_{m+j}\right)$ for every $m$ and every family $g$.

Write $u_j:=g_{m+j}$, a family $\mathbb{N}\to M$, so that ${\prod }_{j<n}{g}_{m+j}={\prod }_{j<n}{u}_j$ and $u_n=g_{m+n}$.

In claim 2 the natural $c_j$ exists because $s_j\le s_{\sigma (j)}$, and it is unique by cancellation, so $B_j$ is well defined for every $j$ and $B:\mathbb{N}\to M$ is a family.

Regrouping law, base $r=0$: the left side is ${\prod }_{j<0}{B}_j=e$ and the right side is ${\prod }_{i<s_0}{g}_i={\prod }_{i<0}{g}_i=e$.

Regrouping law, inductive hypothesis: assume ${\prod }_{j<r}{B}_j={\prod }_{i<s_r}{g}_i$ for this $r$.

Commutative case, base $n=0$: both products are empty, hence both equal $e$.

Commutative case, inductive hypothesis: fix $n$ and assume that for every family $g$, every $\pi \in \mathrm{Sym}(n)$ and every family $h$ with $h_i=g_{\pi (i)}$ for $i<n$, one has ${\prod }_{i<n}{h}_i={\prod }_{i<n}{g}_i$.

Successor step, the data. Let $\pi \in \mathrm{Sym}(\sigma (n))$ and let $h$ satisfy $h_i=g_{\pi (i)}$ for every $i<\sigma (n)$. Put $k:={\pi }^{-1}(n)$, so $k<\sigma (n)$, that is $k\le n$, and let $r$ be the natural with $k+r=n$; then $k+\sigma (r)=\sigma (k+r)=\sigma (n)$.

The skip map. Define $\delta :\mathbb{N}\to \mathbb{N}$ by $\delta (i)=i$ when $i<k$ and $\delta (i)=\sigma (i)$ when $k\le i$; the two clauses are exhaustive and mutually exclusive by trichotomy, so $\delta$ is a function. It is injective: two arguments below $k$ with equal images are equal; two arguments at least $k$ have equal images only if their successors agree, hence only if they agree; and if $i<k\le j$ then $\delta (i)=i<k\le j<\sigma (j)=\delta (j)$.

Splitting law, successor step: $m+\sigma (n)=\sigma (m+n)$, so ${\prod }_{i<m+\sigma (n)}{g}_i={\prod }_{i<\sigma (m+n)}{g}_i=\left({\prod }_{i<m+n}{g}_i\right)\cdot g_{m+n}$; applying the hypothesis and then associativity, this equals $\left(\right({\prod }_{i<m}{g}_i)\cdot {\prod }_{j<n}{u}_j)\cdot u_n=\left({\prod }_{i<m}{g}_i\right)\cdot \left(\right({\prod }_{j<n}{u}_j)\cdot u_n)=\left({\prod }_{i<m}{g}_i\right)\cdot {\prod }_{j<\sigma (n)}{u}_j$, which is the claim at $\sigma (n)$.

$\delta$ maps $n$ into $\sigma (n)\setminus \{k\}$, and onto it. If $i<k$ then $\delta (i)=i\mathrm{\ne }k$ and $i<k\le n<\sigma (n)$. If $k\le i<n$ then $\delta (i)=\sigma (i)$ with $k\le i<\sigma (i)$, so $\delta (i)\mathrm{\ne }k$, and $i<n$ gives $\sigma (i)<\sigma (n)$ by compatibility of the order with adding $1$. Conversely let $y<\sigma (n)$ with $y\mathrm{\ne }k$. If $y<k$ then $y<k\le n$, so $y<n$ and $y=\delta (y)$. If $k<y$ then $y\mathrm{\ne }0$, so $y=\sigma (x)$ for some $x$; from $k<\sigma (x)$ we get $k\le x$, and from $\sigma (x)<\sigma (n)$ we get $x<n$, so $y=\sigma (x)=\delta (x)$ with $x<n$.

By induction the splitting law holds for all $m,n\in \mathbb{N}$; this is claim 1.

The reduced permutation. Set ${\pi }^{\mathrm{\prime }}:=\pi \circ \delta$ restricted to $n$. By step 2.2 it maps $n$ into $\pi [\sigma (n)\setminus \{k\}]$, and since $\pi$ is a bijection of $\sigma (n)$ with $\pi (k)=n$, that image set is $\sigma (n)\setminus \{n\}=n$; the same step gives surjectivity onto it, and ${\pi }^{\mathrm{\prime }}$ is injective as a composite of injections. So ${\pi }^{\mathrm{\prime }}\in \mathrm{Sym}(n)$.

Regrouping law, successor step: ${\prod }_{j<\sigma (r)}{B}_j=\left({\prod }_{j<r}{B}_j\right)\cdot B_r=\left({\prod }_{i<s_r}{g}_i\right)\cdot {\prod }_{l<c_r}{g}_{s_r+l}$ by the hypothesis, and the splitting law with $m=s_r$ and $n=c_r$ turns the right-hand side into ${\prod }_{i<s_r+c_r}{g}_i={\prod }_{i<s_{\sigma (r)}}{g}_i$. By induction claim 2 holds for every $r$.

Splitting the permuted product. By step 1.9 and the splitting law, ${\prod }_{i<\sigma (n)}{h}_i={\prod }_{i<k+\sigma (r)}{h}_i=\left({\prod }_{i<k}{h}_i\right)\cdot \left({\prod }_{j<\sigma (r)}{h}_{k+j}\right)$; and since $\sigma (r)=1+r$, a second application of the splitting law to the family $j\mapsto h_{k+j}$ gives ${\prod }_{j<\sigma (r)}{h}_{k+j}=\left({\prod }_{j<1}{h}_{k+j}\right)\cdot \left({\prod }_{l<r}{h}_{k+1+l}\right)=h_k\cdot {\prod }_{l<r}{h}_{k+\sigma (l)}$.

Recognising the shorter permuted product. Let $w_i:=h_{\delta (i)}$. For $i<k$ one has $w_i=h_i$, and for $l<r$ one has $w_{k+l}=h_{\sigma (k+l)}=h_{k+\sigma (l)}$; so the splitting law at $m=k$, $n=r$ gives ${\prod }_{i<n}{w}_i=\left({\prod }_{i<k}{h}_i\right)\cdot {\prod }_{l<r}{h}_{k+\sigma (l)}$. Moreover $w_i=h_{\delta (i)}=g_{\pi (\delta (i))}=g_{{\pi }^{\mathrm{\prime }}(i)}$ for every $i<n$, by step 2.2 and step 3.2.

Moving the removed factor to the end. $h_k=g_{\pi (k)}=g_n$, and $M$ is commutative, so $\left({\prod }_{i<k}{h}_i\right)\cdot (h_k\cdot {\prod }_{l<r}{h}_{k+\sigma (l)})=\left(\right({\prod }_{i<k}{h}_i)\cdot {\prod }_{l<r}{h}_{k+\sigma (l)})\cdot g_n$, using commutativity to exchange $h_k$ with the block to its right and associativity to regroup.

Applying the hypothesis. The family $w$ satisfies $w_i=g_{{\pi }^{\mathrm{\prime }}(i)}$ for $i<n$ with ${\pi }^{\mathrm{\prime }}\in \mathrm{Sym}(n)$, so the inductive hypothesis gives ${\prod }_{i<n}{w}_i={\prod }_{i<n}{g}_i$; combining with steps 4.2, 5.1 and 4.3, ${\prod }_{i<\sigma (n)}{h}_i=\left({\prod }_{i<n}{g}_i\right)\cdot g_n={\prod }_{i<\sigma (n)}{g}_i$, which is the claim at $\sigma (n)$.

By induction claim 3 holds for every $n$; with claims 1 and 2 already established, the theorem is proved.