$a_0,a_1,\ldots,a_n$, a sequence of integers. Then $\prod\limits_{0\leq iLet $a_0,a_1,\ldots,a_n$ be a sequence of integers.
Then $\prod\limits_{0\leq i<j \leq n} (a_j-a_i)$ is divisible by $\prod\limits_{1\leq i\leq n} i!$.
The same result for only $n!$ is straightforward.
I've tried to do it using induction in $n$ and then on the value of $n$.
 A: This product is the Vandermonde determinant, which strongly suggests the idea of looking for a variation of the Vandermonde matrix. In fact it looks an awful lot like we're supposed to divide the $i^{th}$ column of the Vandermonde matrix by $i!$ or something like that. This suggests the following idea.

Claim: We have


$$\frac{\prod_{0 \le i < j \le n} (a_j - a_i)}{\prod_{1 \le i \le n} i!} = \det \left[ \begin{array}{ccccc} 1 & a_0 & {a_0 \choose 2} & \dots & {a_0 \choose n}\\
 1 & a_1 & {a_1 \choose 2} & \dots & {a_1 \choose n} \\
 1 & a_2 & {a_2 \choose 2} & \dots & {a_2 \choose n}
  \\ 
 \vdots & \vdots & \vdots  & \ddots & \vdots \\
 1 & a_n & {a_n \choose 2} & \dots & {a_n \choose n} \end{array} \right]$$


which is an integer when each $a_i$ is an integer.

Proof. This matrix 1) consists of integer entries so has determinant an integer, and 2) can be obtained from the Vandermonde matrix by first dividing the $i^{th}$ column by $i!$ and then performing column operations which do not change the determinant. So its determinant is the determinant of the Vandermonde matrix divided by $\prod_{1 \le i \le n} i!$ and this is an integer.
Alternatively we can adapt the usual Vandermonde determinant argument: the determinant is zero whenever any two of the $a_i$ are equal, so as a polynomial is divisible by $\prod_{0 \le i < j \le n} (a_j - a_i)$, and by counting its degree it must be a scalar multiple of this product, so we only have to figure out the scalar by evaluating any nonzero example. Then if we set $a_i = i$ the result is a lower triangular matrix with $1$'s on the diagonal, hence the determinant is $1$ in this case. So the determinant is
$$\frac{\prod_{0 \le i < j \le n} (a_j - a_i)}{\prod_{0 \le i < j \le n} (j - i)}$$
which is equivalent to the desired result. $\Box$
