Parity of Inverse of Permutation

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

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

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

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

The result follows immediately.