Transposition/Sign/Fact/Proof3

Suppose that the transposition ${\displaystyle {}\tau }$ swaps the numbers ${\displaystyle {}k<\ell }$. Then

${\displaystyle {}{\begin{aligned}\operatorname {sgn} (\tau )&=\prod _{i<j}{\frac {\tau (j)-\tau (i)}{j-i}}\\&=\prod _{i<j,\,i,j\neq k,\ell }{\frac {\tau (j)-\tau (i)}{j-i}}\cdot \prod _{i<j,\,i=k,j\neq \ell }{\frac {\tau (j)-\tau (i)}{j-i}}\cdot \prod _{i<j,\,j=k}{\frac {\tau (j)-\tau (i)}{j-i}}\cdot \prod _{i<j,\,i\neq k,j=\ell }{\frac {\tau (j)-\tau (i)}{j-i}}\cdot \prod _{i<j,\,i=\ell }{\frac {\tau (j)-\tau (i)}{j-i}}\cdot \prod _{i<j,\,i=k,j=\ell }{\frac {\tau (j)-\tau (i)}{j-i}}\\&=\prod _{j>k,\,j\neq \ell }{\frac {j-\ell }{j-k}}\cdot \prod _{i<k}{\frac {\ell -i}{k-i}}\cdot \prod _{i\neq k,\,i<\ell }{\frac {k-i}{\ell -i}}\cdot \prod _{\ell <j}{\frac {j-k}{j-\ell }}\cdot {\frac {k-\ell }{\ell -k}}\\&=\prod _{k<j<\ell }{\frac {j-\ell }{j-k}}\cdot \prod _{k<i<\ell }{\frac {k-i}{\ell -i}}\cdot (-1)\\&=-1.\end{aligned}}}$

In the penultimate equation, we have combined the first and the fourth and the second and the third product. The last equation follows, since we have the same index set in the two products and they cancel each other.