Proof by induction: For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :If $\circ_1, \ldots, \circ_n$ are all [Definition:Associative Operation|associative]], then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_n$. === Basis for the Induction === $P \left({1}\right)$ is true, as this just says: :$\circ_1$ is associative. $P \left({2}\right)$ is the case: :If $\circ_1$ and $\circ_2$ are both associative, then so is the external direct product $\circ$ of $\circ_1$ and $\circ_2$. This has been proved in External Direct Product Associativity. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :If $\circ_1, \ldots, \circ_k$ are all associative, then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_k$. Then we need to show: :If $\circ_1, \ldots, \circ_{k+1}$ are all associative, then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_{k+1}$. === Induction Step === This is our induction step: Let $a, b, c \in S^{k+1}$: :$a = \left({a_1, a_2, \ldots, a_k, a_{k+1} }\right)$ :$b = \left({b_1, b_2, \ldots, b_k, b_{k+1} }\right)$ :$c = \left({c_1, c_2, \ldots, c_k, c_{k+1} }\right)$ Note that in the below, by abuse of notation, $\circ$ is to be used for two separate operations: :$\left({a_1, a_2, \ldots, a_k, a_{k+1} }\right) \circ \left({b_1, b_2, \ldots, b_k, b_{k+1} }\right)$ and: :$\left({a_1, a_2, \ldots, a_k}\right) \circ \left({b_1, b_2, \ldots, b_k}\right)$ Thus: {{begin-eqn}} {{eqn | l = a \circ \left({b \circ c}\right)       | r = \left({a_1, a_2, \ldots, a_k, a_{k+1} }\right) \circ \left({\left({b_1, b_2, \ldots, b_k, b_{k+1} }\right) \circ \left({c_1, c_2, \ldots, c_k, c_{k+1} }\right)}\right)       | c =  }} {{eqn | r = \left({\left({a_1, a_2, \ldots, a_k}\right), a_{k+1} }\right) \circ \left({\left({\left({b_1, b_2, \ldots, b_k}\right), b_{k+1} }\right) \circ \left({\left({c_1, c_2, \ldots, c_k}\right), c_{k+1} }\right)}\right)       | c = Definition of Ordered Tuple }} {{eqn | r = \left({\left({a_1, a_2, \ldots, a_k}\right) \circ \left({\left({b_1, b_2, \ldots, b_k}\right) \circ \left({c_1, c_2, \ldots, c_k}\right)}\right), a_{k+1} \circ_{k+1} \left({b_{k+1} \circ_{k+1} c_{k+1} }\right)}\right)       | c = Definition of Induced Operation }} {{eqn | r = \left({\left({\left({a_1, a_2, \ldots, a_k}\right) \circ \left({b_1, b_2, \ldots, b_k}\right)}\right) \circ \left({c_1, c_2, \ldots, c_k}\right), a_{k+1} \circ_{k+1} \left({b_{k+1} \circ_{k+1} c_{k+1} }\right)}\right)       | c = Induction Hypothesis }} {{eqn | r = \left({\left({\left({a_1, a_2, \ldots, a_k}\right) \circ \left({b_1, b_2, \ldots, b_k}\right)}\right) \circ \left({c_1, c_2, \ldots, c_k}\right), \left({a_{k+1} \circ_{k+1} b_{k+1} }\right) \circ_{k+1} c_{k+1} }\right)       | c = Basis for the Induction }} {{eqn | r = \left({\left({\left({a_1, a_2, \ldots, a_k}\right), a_{k+1} }\right) \circ \left({\left({b_1, b_2, \ldots, b_k}\right), b_{k+1} }\right)}\right) \circ \left({\left({c_1, c_2, \ldots, c_k}\right), c_{k+1} }\right)       | c = Definition of Induced Operation }} {{eqn | r = \left({\left({a_1, a_2, \ldots, a_k, a_{k+1} }\right) \circ \left({b_1, b_2, \ldots, b_k, b_{k+1} }\right)}\right) \circ \left({c_1, c_2, \ldots, c_k, c_{k+1} }\right)       | c = Definition of Ordered Tuple }} {{eqn | r = \left({a \circ b}\right) \circ c       | c = By Hypothesis }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: For all $n \in \N_{> 0}$: :If $\circ_1, \ldots, \circ_n$ are all associative, then so is the external direct product $\circ$ of $\circ_1, \ldots, \circ_n$. {{qed}}