Permutation/Sign via inversions/Fact/Proof

Proof

We write

${}{\begin{aligned}\operatorname {sgn} (\pi )&=\prod _{i<j}{\frac {\pi (j)-\pi (i)}{j-i}}\\&=\prod _{(i,j)\in F}{\frac {\pi (j)-\pi (i)}{j-i}}\prod _{(i,j)\not \in F}{\frac {\pi (j)-\pi (i)}{j-i}}\\&=(-1)^{k}\prod _{(i,j)\in F}{\frac {\pi (i)-\pi (j)}{j-i}}\prod _{(i,j)\not \in F}{\frac {\pi (j)-\pi (i)}{j-i}}\\&=(-1)^{k},\end{aligned}}$

because, after this reordering, we have in the numerator as well as in the denominator the product of all positive differences.

To fact