Inverse of Group Product/General Result/Proof 2

Theorem
Let $\struct {G, \circ}$ 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:
 * $\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$

Proof
Proof by induction:

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

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

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

which has been proved in Inverse of Group Product.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \paren k$ is true, where $k \ge 2$, then it logically follows that $P \paren {k + 1}$ 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 $P \paren k \implies P \paren {k + 1}$ and the result follows by the Principle of Mathematical Induction.

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