External Direct Product Associativity/General Result

Theorem
Let $\displaystyle \left({S, \circ}\right) = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\left({S_1, \circ_1}\right), \left({S_2, \circ_2}\right), \ldots, \left({S_n, \circ_n}\right)$.

If $\circ_1, \ldots, \circ_n$ are all associative, then so is $\circ$.

Proof
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 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:

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 $\circ_1, \ldots, \circ_n$.