Inverse of Group Product/General Result/Proof 1

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $a_1, a_2, \ldots, a_n \in G$, 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
We have that a group is a monoid, all of whose elements are invertible.

The result follows from Inverse of Product/Monoid/General Result.