Parity of Inverse of Permutation

Theorem
Let $S_n$ denote the symmetric group on $n$ letters.

Then $\forall \pi \in S_n: \operatorname{sgn} \left({\pi}\right) = \operatorname{sgn} \left({\pi^{-1}}\right)$.

Proof
From Parity of a Permutation, $\operatorname{sgn} \left({I_{S_n}}\right) = 1$.

Thus $\pi \pi^{-1} = I_{S_n} \implies \operatorname{sgn} \left({\pi}\right) \operatorname{sgn} \left({\pi^{-1}}\right) = 1$.

The result follows immediately.