Inverse of Product/Monoid/General Result

Theorem
Let $\left({S, \circ}\right)$ 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 $\left({a_1 \circ a_2 \circ \cdots \circ a_n}\right)^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$.

Proof
Proof by induction:

We have, from above, $\left({a_1 \circ a_2}\right)^{-1} = a_2^{-1} \circ a_1^{-1}$, and (trivially) $\left({a_1}\right)^{-1} = a_1^{-1}$.

Assume that $\left({a_1 \circ a_2 \circ \cdots \circ a_k}\right)^{-1} = a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$.

Then:

So the assumption being true for $n = k$ implies that it is also true for $n = k + 1$.

It is also true for $n = 1$ (trivially) and $n = 2$ (proved above).

So, by the Principle of Mathematical Induction, it is true for all $n \in \N^*$.