Verify that for any $n\times n$ Matrix $A$ we have: $\det A^t=\det A$.
$a_{\pi(1)1}\cdots a_{\pi(n)n}=a_{1\pi^{-1}(1)}\cdots a_{n\pi^{-1}(n)}$ and $\sign(\pi)=\sign(\pi^{-1})$.