Friend's Beautiful Problem

Note: Every time I use the vertical bar $|$ , it will mean "divides".

Let $p$ be a prime, let $n$ be any positive integer, and let $\alpha$ be any non-negative integer. Then $p^\alpha|n! \implies \alpha \leq \sum_{k=1}^\infty \left\lfloor\frac{n}{p^k}\right\rfloor \leq \sum_{k=1}^\infty \frac{n}{p^k} \leq \sum_{k=1}^\infty \frac{n}{2^k} = n.$ Therefore $n!$ divides $\displaystyle \prod_{\substack{p \leq n \\ p \text{ prime}}} p^n$

Now consider $x$ and $m$ as in the problem. For any prime $p \leq m$ , we have $p | m! | x^m \implies p | x^m \implies p | x$ Therefore, if $n < m$ , then $n! \Big| \prod_{\substack{p \leq n \\ p \text{ prime}}} p^n \Big| \prod_{\substack{p \leq m \\ p \text{ prime}}} p^n \Big| x^n \implies n! | x^n \qquad \qquad \square$