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> Show that the product of any $m$ consecutive positive integers is divisible by $m!$.
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Note that we have that $\frac{n!}{m!(n-m)!} \in \mathbb{Z}$ for $0 \leq m \leq n$.

$\require{cancel} \frac{n!}{m!(n-m)!} = \frac{n\cdot(n-1)\cdot\cdot\cdot(n-m+1)(n-m)!}{m!(n-m)!} = \frac{n\cdot(n-1)\cdot\cdot\cdot(n-m+1)\cancel{(n-m)!}}{m!\cancel{(n-m)!}} = \frac{n\cdot(n-1)\cdot\cdot\cdot(n-m+1)}{m!}$ where $n\cdot(n-1)\cdot\cdot\cdot(n-m+1)$ are $m$ consecutive integers divisible by $m!$

Of course, I'm using the fact that I know the combinations formula is an integer but given that, is the proof this simple? Or did I make an error here. Appreciate help!