Inverse Element is Power of Order Less 1

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Theorem

Let $G$ be a group whose identity is $e$.

Let $g \in G$ be of finite order.

Then:

$\left\vert{g}\right\vert = n \implies g^{n-1} = g^{-1}$

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \left\vert{g}\right\vert$	$=$	$\displaystyle n$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle g^n$	$=$	$\displaystyle e$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle g^n g^{-1}$	$=$	$\displaystyle e g^{-1}$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle g^{n - 1}$	$=$	$\displaystyle g^{-1}$	$\displaystyle $	$\displaystyle $	$\displaystyle $

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{g}\right\vert\)	\(=\)	\(\displaystyle n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle g^n\)	\(=\)	\(\displaystyle e\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle g^n g^{-1}\)	\(=\)	\(\displaystyle e g^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle g^{n - 1}\)	\(=\)	\(\displaystyle g^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)