Inverse of Product/Monoid/General Result

Theorem
Let $\struct {S, \circ}$ be a monoid whose identity is $e$.

Let $a_1, a_2, \ldots, a_n \in S$ be invertible for $\circ$, with inverses ${a_1}^{-1}, {a_2}^{-1}, \ldots, {a_n}^{-1}$.

Then $a_1 \circ a_2 \circ \cdots \circ a_n$ is invertible for $\circ$, and:
 * $\forall n \in \N_{> 0}: \paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = {a_n}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
 * $\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = {a_n}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$

$\map P 1$ is (trivially) true, as this just says:
 * $\paren {a_1}^{-1} = {a_1}^{-1}$

Basis for the Induction
$\map P 2$ is the case:
 * $\paren {a_1 \circ a_2}^{-1} = {a_2}^{-1} \circ {a_1}^{-1}$

which has been proved in Inverse of Product in Monoid.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\paren {a_1 \circ a_2 \circ \cdots \circ a_k}^{-1} = {a_k}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$

Then we need to show:
 * $\paren {a_1 \circ a_2 \circ \cdots \circ a_{k + 1} }^{-1} = {a_{k + 1} }^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N_{> 0}: \paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = {a_n}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$