Let $P$ be the additive group of mappings from Z$^{[1,n]}$ to Z and let $\mathfrak{S}_n$ denote the symmetric group of the interval $[1,n]$ of the set N of natural numbers.
For $\sigma\in\mathfrak{S}_n$ and $f\in P$, let $\sigma f$ be the element of $P$ defined by $$\sigma f(z_1,...,z_n)=f(z_{\sigma (1)},...,z_{\sigma (n)}).$$
This defines a left-operation of $\mathfrak{S}_n$ on $P$ such that $\sigma(- f)=-\sigma f$ for all $\sigma\in\mathfrak{S}_n$ and $f\in P.$
Now, let $p$ be the element of $P$ defined by
$$p(z_1,....z_n)=\prod_{i<j}(z_j-z_i).$$
Claim. For $\sigma\in\mathfrak{S}_n, (-1)^{\upsilon(\sigma)}p=\sigma p$, where $\upsilon(\sigma)$ denotes the number of inversions of $\sigma$.
Proof. Let $C$ be the set of ordered pairs $(i,j)$ such that $1\leq i\leq n$, $1\leq j \leq n$, and $i<j$. A permutation $\theta$ is defined one $C$ by setting $\theta(i,j)=(\sigma(i),\sigma(j))$ if $(i,j)$ is not an inversion, and $\theta(i,j)=(\sigma(j),\sigma(i))$ if $(i,j)$ is an inversion. This implies $(-1)^{\upsilon(\sigma)}p=\sigma p$.
Why does the existence of $\theta$ imply the equality $(-1)^{\upsilon(\sigma)}p=\sigma p$?
To begin with, we have
$$\sigma p(z_1,....z_n)=(z_{\sigma (1)},...,z_{\sigma (n)})=\prod_{i<j}(z_{\sigma (j)}-z_{\sigma (i)});$$
What is the next step (algebraically I mean), given $\theta$?
Another approach. Let $\tau$ be a transposition. Then $\tau p=-p$. Recall that every permutation $\sigma$ can be written as a product of transpositions $\sigma=\tau_1\cdots\tau_v$ where $v\equiv v(\sigma)\pmod 2$. Then $$\sigma p=(\tau_1\cdots\tau_v)p=(-1)^vp$$