Prove the sign of $\sigma$ equals $\prod_{1\leq i < j \leq n} \dfrac{\sigma(j)-\sigma(i)}{j-i}.$ 
How can one prove that for a permutation $\sigma$ of $\{1,\cdots, n\}$, the sign of $\sigma$ equals $\prod_{1\leq i < j \leq n} \dfrac{\sigma(j)-\sigma(i)}{j-i}?$ The sign of a permutation is the parity of the number of transpositions it can be written as.

Let ${\rm Inv}(\sigma)$ denote the number of inversions of $\sigma$.
From this post, $(-1)^{{\rm Inv}(\sigma)} = \prod_{1\leq i < j \leq n} \dfrac{\sigma(j)-\sigma(i)}{j-i},$ so one method is to prove that the sign equals $(-1)^{{\rm Inv}(\sigma)}$. Though I think there's a more direct approach that doesn't use the equality given by that formula. Also, I know that all permutations have a representation as a product of disjoint cycles and that every permutation is either even or odd but not both. I'm not sure if the claim can be proven by induction on $n$, the number of elements the permutation is acting on.
 A: Let's define $\varphi : \sigma \mapsto \displaystyle\prod_{1\leq i < j \leq n} \dfrac{\sigma(j)-\sigma(i)}{j-i}$.

*

*We prove that $\varphi(\sigma \circ \tau)=\varphi(\sigma) \varphi(\tau)$. Indeed,
\begin{align*} \varphi(\sigma \circ \tau) & =  \prod_{1\leq i < j \leq n} \dfrac{\sigma \circ \tau(j)-\sigma\circ \tau(i)}{j-i}\\& =  \prod_{1\leq i < j \leq n} \dfrac{\sigma \circ \tau(j)-\sigma\circ \tau(i)}{\tau(j)-\tau(i)} \times \dfrac{\tau(j)-\tau(i)}{j-i}\\& =  \prod_{1\leq i < j \leq n} \dfrac{\sigma \circ \tau(j)-\sigma\circ \tau(i)}{\tau(j)-\tau(i)} \times \prod_{1\leq i < j \leq n}\dfrac{\tau(j)-\tau(i)}{j-i}\\& =  \prod_{1\leq i < j \leq n} \dfrac{\sigma (j)-\sigma(i)}{j-i} \times \prod_{1\leq i < j \leq n}\dfrac{\tau(j)-\tau(i)}{j-i}\\&=\varphi(\sigma)\varphi(\tau)
 \end{align*}


*We prove that if $\tau$ is a transposition, then $\varphi(\tau)=-1$. Indeed, if $\tau$ is the $(i,j)$ transposition, then
$$\varphi(\tau)= -\prod_{k \neq i,j}\dfrac{\tau(k)-\tau(i)}{k-i}\dfrac{\tau(k)-\tau(j)}{k-j} =-\prod_{k \neq i,j}\dfrac{k-j}{k-i}\dfrac{k-i}{k-j} = -1 $$


*The result directly follows from the two points above.
A: Consider the polynomial in $n$ variables given by  $\prod_{1\leq i < j \leq n} (x_j-x_i)$ and the obvious action that $S_n$ has on the indices of those variables. It is not hard to see that a transposition acts on that polynomial by changing the sign since $x_i-x_j=-(x_j-x_i)$. This means for some permutation $\sigma$ we have that $$(-1)^{\rm Inv {(\sigma})}\displaystyle\prod_{1\leq i < j \leq n} (x_j-x_i) = \displaystyle\prod_{1\leq i < j \leq n} (x_{\sigma(j)}-x_{\sigma(i)})$$ which gives us that $$(-1)^{\rm Inv {(\sigma})}=\displaystyle\prod_{1\leq i < j \leq n} \frac{x_{\sigma(j)}-x_{\sigma(i)}}{x_j-x_i}$$
and the result follows as a corollary.
