Let $I_n:=\{1, \ldots, n\}$ where $n\in\mathbb N$. I'm wondering whether is there a way to write $\displaystyle\prod_{i<j} (j-i)$ as a sum?
The product above is taken over all pairs $(i, j)\in I_n\times I_n$ such that $i<j$.
Thanks.
Context: I want to show $$\prod_{i<j} (\sigma_j-\sigma_i)=(-1)^{\imath(\sigma)}\prod_{i<j} (j-i),$$ where $\sigma$ is a permutation of $I_n$, $\sigma_j:=\sigma(j)$ and $$\imath(\sigma):=\#\{(i, j)\in I_n\times I_n; i<j\quad \textrm{and}\ \sigma(i)>\sigma(j)\},$$ that is $\imath(\sigma)$ is the number of iversions $\sigma$ performs.
It can be proven by mathematical induction that for a set $I_n:=\{1, \ldots, n\}$, the set of distances between pairs of its elements is $D_n:=\{n, n-1, \ldots, 2, 1\}$, where the first element represents the number of pairs with distance $1$, and the last element is the number of pairs with distance $n$. This is equivalent to $D_n=n-I_n+1$. Your product is asking for the $D_n(k)$ multiples of $k$, or, in other words, $$\displaystyle\prod_{k=1}^{n} (n-k+1)k=n(n-1)!\cdot n!=(n!)^2$$ While not a sum, I think this result is pretty important.