Inverse of Group Product/General Result/Proof 2

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
Proof by induction:

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

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

Basis for the Induction
$P \left({2}\right)$ is the case:
 * $\left({a_1 \circ a_2}\right)^{-1} = a_2^{-1} \circ a_1^{-1}$

which has been proved in Inverse of Group Product.

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:
 * $\left({a_1 \circ a_2 \circ \cdots \circ a_k}\right)^{-1} = a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$

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

Induction Step
This is our induction step:

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

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