As in the title stated I want to simplify $\frac{\prod_{1\leq j <i \leq n}(i-j)(i-j)}{\prod_{1\leq i,j\leq n}(i+j-1)}$. It occurs as the determinant of the Hilbert matrix after using the formula for the determinant of a Cauchy matrix. However, I am not sure how to simplify such a product. My efforts led me to
$$\prod_{1\leq j < i \leq n}(i-j)=\prod_{i=1}^{n-1}i! $$ and $$\prod_{1\leq i,j\leq n}(i+j-1)=\prod_{ i=1}^{2n-1}i!.$$
I used the wrong entries for the matrix. The right fraction expression is $$\frac{\prod_{1\leq j <i \leq n}(i-j)(i-j)}{\prod_{1\leq i,j\leq n}(i+j-1)}=\frac{\prod_{1\leq j <i \leq n}(i-j)^2}{\prod_{1\leq i,j\leq n}(i+j-1)}=\frac{\prod_{i=1}^{n-1}i!}{\prod_{i=n}^{2n-1}i!}$$ However, it does not give me the right determinant, so there has to be a mistake somewhere
if $n\geq 3$ in the $[1,n]$ you can find a pair $(j<i)$ such that $i-j=2$
in this case is $0$
then you can calculate for n=1,2