The proof will proceed by the Principle of Mathematical Induction on $\N$. Let $T$ be the set of all $n \in \N_{>0}$ such that: :$(1): \quad$ for every sequence $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + n}$ of elements of $S$ and: :$(2): \quad$ for every strictly increasing sequence $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ of natural numbers such that $r_0 = p$ and $r_s = p + n$: the statement: :$\displaystyle b_k = \prod_{j \mathop = r_{k - 1} \mathop + 1}^{r_k} a_j$ holds. === Basis for the Induction === Let $n = 1$. Then: {{begin-eqn}} {{eqn | l = n       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = r_s       | r = r_0 + 1       | c =  }} {{eqn | ll= \leadsto       | l = s       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \prod_{k \mathop = 1}^s b_k       | r = b_1       | c =  }} {{eqn | r = a_{p + 1}       | c =  }} {{eqn | r = \prod_{k \mathop = p + 1}^{p \mathop + n} a_k       | c =  }} {{end-eqn}} So $1 \in T$. 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: :$(1): \quad$ for every sequence $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + m}$ of elements of $S$ and: :$(2): \quad$ for every strictly increasing sequence $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ of natural numbers such that $r_0 = p$ and $r_s = p + m$: the statement: :$\displaystyle b_k = \prod_{j \mathop = r_{k-1} \mathop + 1}^{r_k} a_j$ holds. It is to be demonstrated that it follows that: :$(1): \quad$ for every sequence $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + m + 1}$ of elements of $S$ and: :$(2): \quad$ for every strictly increasing sequence $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ of natural numbers such that $r_0 = p$ and $r_s = p + m + 1$: the statement: :$\displaystyle b_k = \prod_{j \mathop = r_{k - 1} \mathop + 1}^{r_k} a_j$ holds. === Induction Step === This is our induction step: Let $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + m + 1}$ be a sequence of elements of $S$. Let $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ be a strictly increasing sequence of natural numbers such that $r_0 = p$ and $r_s = p + m + 1$. Then $r_{s - 1} \le p + m$. There are two cases: :$(1): \quad r_{s - 1} = p + m$ :$(2): \quad r_{s - 1} < p + m$ First, suppose: :$r_{s - 1} = p + m$ Then: :$b_s = a_{p + m + 1}$ Thus: {{begin-eqn}} {{eqn | l = a_{p + 1} \circ \ldots \circ a_{p + m}       | r = b_1 \circ \ldots \circ b_{s-1}       | c = $m \in T$ by Induction Hypothesis }} {{eqn | ll=\leadsto       | l = a_{p + 1} \circ \ldots \circ a_{p + m + 1}       | r = \paren {a_{p + 1} \circ \ldots \circ a_{p + m} } \circ a_{p + m + 1}       | c = {{Defof|Composite (Abstract Algebra)}} }} {{eqn | r = \paren {b_1 \circ \ldots \circ b_{s - 1} } \circ b_s       | c =  }} {{eqn | r = b_1 \circ \ldots \circ b_s       | c = {{Defof|Composite (Abstract Algebra)}} }} {{end-eqn}} Secondly, suppose: :$r_{s - 1} < p + m$ Let $b\,'_s = a_{r_{s - 1} + 1} \circ \ldots \circ a_{r_s + 1}$. Then by definition of composite: :$b_s = b\,'_s \circ a_{p + m + 1}$ Now by the Induction Hypothesis: :$m \in T \implies a_{p + 1} \circ \ldots \circ a_{p + m} = b_1 \circ \ldots \circ b_{s - 1} \circ b\,'_s$ Thus: {{begin-eqn}} {{eqn | l = b_1 \circ \ldots \circ b_s       | r = \paren {b_1 \circ \ldots \circ b_{s - 1} } \circ \paren {b\,'_s \circ a_{p + m + 1} }       | c =  }} {{eqn | r = \paren {\paren {b_1 \circ \ldots \circ b_{s - 1} } \circ b\,'_s} \circ a_{p + m + 1}       | c =  }} {{eqn | r = \paren {b_1 \circ \ldots \circ b_{s - 1} \circ b\,'_s} \circ a_{p + m + 1}       | c =  }} {{eqn | r = \paren {a_{p + 1} \circ \ldots \circ a_{p + m} } \circ a_{p + m + 1}       | c =  }} {{eqn | r = a_{p + 1} \circ \ldots \circ a_{p + m} \circ a_{p + m + 1}       | c =  }} {{end-eqn}} Thus in both cases $m + 1 \in T$. So by the Principle of Mathematical Induction: :$T = \N_{>0}$ Hence the result. {{Qed}}