Let $a_1<a_2<\ldots<a_{n+1}$ different integers. Prove that $$n!\Big|\prod_{1\leq i< j\leq n+1}{(a_j-a_i)}$$Here's my proof:
Let $k\in\bigl\{1,2,\ldots,n\bigr\}$. $k<n+1$ and using the Pigeonhole Principle, there are $0\leq i<j\leq n+1$ (but $0\leq i<j\leq k+1$ is enough) such that $$a_i \equiv a_j\pmod{k}$$Thus, $$k\big|(a_j-a_i)$$Every $k\in\bigl\{1,2,\ldots,n\bigr\}$ satisfies this, and we are done.
Is the proof correct? I couldn't find a mistake myself but it seemed weird that the use of the Pigeonhole Principle had $n+1$ pigeons and only $k$ holes (when $k$ can be as little as $1,2,3$). Am I missing something?
It seems to me that your argument shows that $$ \prod_{1\leq i<j\leq n+1}(a_j-a_i) $$ (you shouldn't allow $i=j$), which is actually a Vandermonde determinant, is divisible by all $k\leq n$.
This is not (yet) enough to conclude that $n!$ divides it: $3\cdot4\cdot 5=60$ has this property but is actually smaller than $5!$.
But it is possible that the argument can be improved upon.
Given a prime $p\leq n<n+1$ there are always $\lfloor\frac n{p^k}\rfloor+1$ elements among the $a_i$'s that give the same class modulo $p^k$ as long as $p^k\leq n$. This should account for the $$ \lfloor\frac n{p}\rfloor+\lfloor\frac n{p^2}\rfloor+\lfloor\frac n{p^3}\rfloor+\cdots $$ factors of $p$ that occur in $n!$.
(I leave to you to fill the details)