Inspired by this I decided to show this:
Let $P_n=\displaystyle\prod_{1\leq i<j\leq n} \big(x_i-x_j\big)$ where $x_1\,\dots\, x_n$ are arbitrary integers. Prove $n!\,\big|\, 2P_n$.
I am wondering whether this can be done using a slick, group theoretic argument.
Ideas:
If $X$ is the set of polynomials in $n$ variables, we can consider $S_n$ acting on $X$ likewise:
$$\pi \Big( p(x_1,\,\dots,\,x_n)\Big)=p\Big(x_{\pi (1)},\,\dots,\,x_{\pi (n)}\Big)$$
Then $\text{stab} \big(P_n\big)=A_n$ so $n!$ divides $2\big|\,\text{stab} \big(P_n\big)\big|$. I'd like to think this can be used to derive the result, but I may be wrong. If not on the other hand, a hint would be great.
This does not seem to be true in general. Look at $\{5,4,3,2,1\}$ Then $$P_n = (5-4)(5-3)(5-2)(5-1)(4-3)...(3-1)(2-1)=4!3!2!1!$$ In this case $2P_n$ is not divisible by $5!$, as $2P_n$ is not divisible by $5$.
A similar case can happen whenever $n$ is prime.