Problem: Prove that for any positive integer $k$,
\[ (k^2)!\cdot\displaystyle\prod_{j=0}^{k-1}\frac{j!}{(j+k)!} \]
is an integer.

Solution: We show the exponent of any given prime $p$ is nonnegative in the expression.
Recall that the exponent of $p$ in $n!$ is equal to
$\sum_{i \ge 1} \left\lfloor n/p^i \right\rfloor$.
In light of this, it suffices to show
that for any prime power $q$, we have
\[
\left\lfloor \frac{k^2}{q} \right\rfloor
+ \sum_{j=0}^{k-1} \left\lfloor \frac{j}{q} \right\rfloor
\ge \sum_{j=0}^{k-1} \left\lfloor \frac{j+k}{q} \right\rfloor
\]
Since both sides are integers, we show
\[
\left\lfloor \frac{k^2}{q} \right\rfloor
+ \sum_{j=0}^{k-1} \left\lfloor \frac{j}{q} \right\rfloor
> -1 + \sum_{j=0}^{k-1} \left\lfloor \frac{j+k}{q} \right\rfloor.
\]
If we denote by $\left\{ x \right\}$ the fractional part of $x$,
then $\left\lfloor x \right\rfloor = x - \left\{ x \right\}$
so it's equivalent to
\[
\left\{ \frac{k^2}{q} \right\}
+ \sum_{j=0}^{k-1} \left\{ \frac{j}{q} \right\}
< 1 + \sum_{j=0}^{k-1} \left\{ \frac{j+k}{q} \right\}.
\]
However, the sum of remainders when $0, 1, \dots, k-1$ are taken modulo $q$
is easily seen to be less than the sum of remainders
when $k, k+1, \dots, 2k-1$ are taken modulo $q$.
So \[ \sum_{j=0}^{k-1} \left\{ \frac{j}{q} \right\}
\le \sum_{j=0}^{k-1} \left\{ \frac{j+k}{q} \right\} \] follows,
and we are done upon noting $\left\{ k^2/q \right\} < 1$.