How prove this $\prod_{1\le i<j\le n}\frac{a_{j}-a_{i}}{j-i}$ is integer

let $a_{i},i=1,2,\cdots,n$ be postive integer ,show that
$$1^{n-1}2^{n-2}\cdots (n-2)^2(n-1)|\prod_{1\le i<j\le n}(a_{i}-a_{j})$$

I know this $\prod_{1\le i<j\le n}(a_{i}-a_{j})$ is
Vandermonde determinants,and I found
$$1^{n-1}2^{n-2}\cdots (n-2)^2\cdot (n-1)=1!2!3!\cdots (n-1)!=\prod_{1\le i<j\le n}(j-i)$$
we only prove $$\prod_{1\le i<j\le n}\dfrac{a_{j}-a_{i}}{j-i}$$ is integer

maybe consider Vandermonde determinants ?
But I can’t prove this

Solutions Collecting From Web of "How prove this $\prod_{1\le i<j\le n}\frac{a_{j}-a_{i}}{j-i}$ is integer"

There are many ways to prove that this number is an integer. For example, it has representation-theoretic and combinatorial interpretations.

Your idea also leads to a solution. We want to prove that $\det\bigl(\frac{a_i^{j-1}}{(j-1)!}\bigr)$ is an integer. Idea: while $x^k/k!$ is not always an integer, there is a deformation
\frac{x^{\downarrow k}}{k!}:=\frac{x(x-1)\ldots(x-k+1)}{k!}=\binom xk
which is always an integer.

Now $\det(a_i^{j-1})=\det(a_i^{\downarrow j-1})$ (in general, if each $P_k$ is a polynomial with leading term $x^k$ then $\det P_{j-1}(x_i)=\det x_i^{j-1}$). So $\det\bigl(\frac{a_i^{j-1}}{(j-1)!}\bigr)=\det\left(\binom{a_i}{j-1}\right)$ which is manifestly an integer.