The proof will proceed by the Principle of Mathematical Induction on $\N_{>0}$. Let $T$ be the set of all $n \in \N_{>0}$ such that: :$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma n} = a_1 \circ \cdots \circ a_n$ holds for all sequences $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ of $n$ elements of $S$ which satisfy: :$\forall i, j \in \closedint 1 n: a_i \circ a_j = a_j \circ a_i$ for every permutation $\sigma: \N_n \to \N_n$. === Basis for the Induction === We have that when $n = 1$: :$\map \sigma 1 = 1$ and so: :$a_{\map \sigma 1} = a_1$ for all $\sigma: \set 1 \to \set 1$. Thus $1 \in T$. This is our basis for the induction. === Induction Hypothesis === It is to be shown that, if $m \in T$ where $m \ge 1$, then it follows that $m + 1 \in T$. This is the induction hypothesis: :$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma m} = a_1 \circ \cdots \circ a_m$ holds for all sequences $\sequence {a_k}_{1 \mathop \le k \mathop \le m}$ of $m$ elements of $S$ which satisfy: :$\forall i, j \in \closedint 1 m: a_i \circ a_j = a_j \circ a_i$ for every permutation $\sigma: \N^*_m \to \N^*_m$. It is to be demonstrated that it follows that: :$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma {m + 1} } = a_1 \circ \cdots \circ a_{m + 1}$ holds for all sequences $\sequence {a_k}_{1 \mathop \le k \mathop \le m + 1}$ of $m + 1$ elements of $S$ which satisfy: :$\forall i, j \in \closedint 1 {m + 1}: a_i \circ a_j = a_j \circ a_i$ for every permutation $\sigma: \N_{m + 1} \to \N_{m + 1}$. === Induction Step === This is our induction step: Suppose $m \in T$. Let $\sequence {a_k}_{1 \le k \le m + 1}$ be a sequence of $m+1$ elements of $S$ which satisfy: :$\forall i, j \in \closedint 1 {n + 1}: a_i \circ a_j = a_j \circ a_i$ Let $\sigma: \N_{m + 1} \to \N_{m + 1}$ be a permutation of $\closedint {1 {m + 1}$. There are three cases to consider: :$(1): \quad \map \sigma {m + 1} = m + 1$ :$(2): \quad \map \sigma 1 = m + 1$ :$(3): \quad \map \sigma r = m + 1$ for some $r \in \closedint 2 m$ $(1): \quad$ Suppose $\map \sigma {m + 1} = m + 1$. Then the restriction of $\sigma$ to $\N_m$ is then a permutation of $\N_m$. From the induction hypothesis: :$m \in T$: Thus: :$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma m} = a_1 \circ \cdots \circ a_m$ from which: {{begin-eqn}} {{eqn | l = a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma {m + 1} }       | r = \paren {a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma m} } \circ a_{\map \sigma {m + 1} }       | c =  }} {{eqn | r = \paren {a_1 \circ \cdots \circ a_m} \circ a_{m + 1}       | c =  }} {{eqn | r = a_1 \circ \cdots \circ a_{m + 1}       | c =  }} {{end-eqn}} $(2): \quad$ Suppose $\map \sigma 1 = m + 1$. Let $\tau: \N^*_m \to \N^*_m$ be the mapping defined as: :$\forall k \in \closedint 1 m: \map \tau k = \map \sigma {k + 1}$ From Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor: :$\closedint 1 {m + 1} = \closedint 1 m \cup \set {m + 1}$ Thus $\tau$ is clearly a permutation on $\closedint 1 m$. Thus, as $m \in T$: :$a_{\map \tau 1} \circ \cdots \circ a_{\map \tau m} = a_1 \circ \cdots \circ a_m$ So: {{begin-eqn}} {{eqn | l = a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma {m + 1} }       | r = a_{\map \sigma 1} \circ \paren {a_{\map \sigma 2} \circ \cdots \circ a_{\map \sigma {m + 1} } }       | c =  }} {{eqn | r = a_{m + 1} \circ \paren {a_{\map \tau 1} \circ \cdots \circ a_{\map \tau {m + 1} } }       | c =  }} {{eqn | r = a_{m + 1} \circ \paren {a_1 \circ \cdots \circ a_m}       | c =  }} {{eqn | r = \paren {a_1 \circ \cdots \circ a_m} \circ a_{m + 1}       | c = Element Commutes with Product of Commuting Elements }} {{eqn | r = a_1 \circ \cdots \circ a_{m + 1}       | c =  }} {{end-eqn}} $(3): \quad$ Suppose $\map \sigma r = m + 1$ for some $r \in \closedint 2 m$. Let $\tau: \N_{m + 1} \to \N_{m + 1}$ be a mapping defined by: :$\forall k \in \N_{m + 1}: \map \tau k = \begin{cases} \map \sigma k & : k \in \closedint 1 {r - 1} \\ \map \sigma {k + 1} & : k \in \closedint r m \\ m + 1 & : k = m + 1 \end{cases}$ It is seen that $\tau$ is a permutation of $\N_{m + 1}$. So, by the first case: :$a_{\map \tau 1} \circ \cdots \circ a_{\map \tau {m + 1} } = a_1 \circ \cdots \circ a_{m + 1}$ Thus: {{begin-eqn}} {{eqn | l = a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma {m + 1} }       | r = \paren {a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma {r - 1} } } \circ \paren {a_{\map \sigma r} \circ \paren {a_{\map \sigma {r + 1} } \circ \cdots \circ a_{\map \sigma {m + 1} } } }       | c =  }} {{eqn | r = \paren {a_{\map \tau 1} \circ \cdots \circ a_{\map \tau {r - 1} } } \circ \paren {a_{\map \tau {m + 1} } \circ \paren {a_{\map \tau r} \circ \cdots \circ a_{\map \tau m} } }       | c =  }} {{eqn | r = \paren {a_{\map \tau 1} \circ \cdots \circ a_{\map \tau {r - 1} } } \circ \paren {\paren {a_{\map \tau r} \circ \cdots \circ a_{\map \tau m} } \circ a_{\map \tau {m + 1} } }       | c =  }} {{eqn | r = a_{\map \tau 1} \circ \cdots \circ a_{\map \tau {m + 1} }       | c =  }} {{eqn | r = a_1 \circ \cdots \circ a_{m + 1}       | c =  }} {{end-eqn}} So in all cases, $m + 1 \in T$. Thus by the Principle of Mathematical Induction: :$T = \N_{>0}$ The result follows. {{qed}}