It is known that for every $n$ consecutive integers, their product is divisible by $n!$, since $${{m}\choose{n}} = \frac{m!}{n!(m-n)!}$$ is also an integer.
So is it true that for every distinct integer $a_1, a_2, ..., a_n$, the number $$S = \prod_{1\leq i< j\leq n} \frac{a_i - a_j}{i-j}$$ is also an integer?
(Sorry for my grammar mistakes, English is my second language)
On
We observe that $\prod_{1 \leq i < j \leq n} (i-j) = \prod_{r=1}^n r!$ and the numerator is just the Vandermonde Determinant so that by performing row operations we obtain $$\prod_{1 \leq i < j \leq n} \frac{a_i-a_j}{i-j} = \begin{vmatrix} 1 & 1 & \cdots & 1 \\ \binom{a_1}{1} & \binom{a_2}{1} & \cdots & \binom{a_n}{1} \\ \binom{a_1}{2} & \binom{a_2}{2} & \cdots & \binom{a_n}{2} \\ \cdots & \cdots & \cdots & \cdots \\ \binom{a_1}{n-1} & \binom{a_2}{n-1} & \cdots & \binom{a_n}{n-1} \\ \end{vmatrix}$$
which is obviously an integer.
Let $q = p^k$ for some prime $p$ and integer $k>0$. If you let $n_q$ = the number of pairs $i<j$ for which $a_i - a_j$ is divisible by $q$, then notice that $n_q$ is minimized for $a_1,\ldots,a_n = 1,2,\ldots,n$ (it is a bit of work to show this but it is not deep) (basically, you consider the set $\mathbb{Z}/(q)$ and now you look the function $f: \mathbb{Z}/(q) \rightarrow \mathbb{N}$ for which $f(a)$ is the number of $i$ for which $a_i$ mod $q$ is $a$. Then observe that you can compute $n_q$ from $f$, and that $n_q$ is minimized if all values of $f$ are inside $\{m,m+1\}$ where $m$ is $n/q$ rounded down).
Now apply this for $q = p, p^2, p^3, \ldots$ and you see that the $p$-adic valuation of the numerator of $S$ is at least that of the denominator. That means $S$ is an integer.