Consider the map $\epsilon(\sigma) := \prod_{i<j}\frac{(\sigma(j)-\sigma(i))}{j-i}$ for $\sigma \in S_n$. Now we are to show that (1) this map is onto $\{-1, 1\}$, that's fairly easy and that (2) the map is a homomorphism. How do we do that? I can't see a way to prove that $\prod_{i<j}\frac{\sigma(\pi(j)) - \sigma(\pi(i))}{j-i} = \prod_{i<j}\frac{(\sigma(j)-\sigma(i))}{j-i} .\prod_{i<j}\frac{(\pi(j)-\pi(i))}{j-i}$ .
I know that $\prod_{i<j} (\sigma(j)-\sigma(i)) = \mathrm{det}(V)$ is the determinant of the Vandermonde matrix with entries $V_{i,j} = \sigma(i)^{j-1}$, but I can't really see how that helps either (supposing it is helpful at all)...
The exercise finishes with demanding to prove that $\mathrm{ker} (\epsilon) = A_n$.
Help is very appreciated!
One way to prove it is the following : $\prod_{i<j}\frac{\sigma(\pi(j))-\sigma(\pi(i))}{j-i} = \prod_{i<j} \frac{\sigma(\pi(j))-\sigma(\pi(i))}{\pi(j)-\pi(i)}\frac{\pi(j)-\pi(i)}{j-i}$
And then you notice that in $\prod_{i<j} \frac{\sigma(\pi(j))-\sigma(\pi(i))}{\pi(j)-\pi(i)}$, if you permute the indices, it's just $\prod_{i<j} \frac{\sigma(j)-\sigma(i)}{j-i}$, which proves that it's a morphism.