On the products of $n$ consecutive numbers

Choose $n$ consecutive whole numbers: $k+1, k+2, \dots, k+n.$ The product of these numbers is always a multiple of $n!$ Prove.

Example: $n=5,$ $(k\mod 5)$ can be $1, 2, 3, 4, 0.$ For $(k\mod 5)=0,$ we have the following numbers $5p+1, 5p+2, 5p+3, 5p+4, 5p+5.$ This ensures that there is one multiple of every number from $1$ to $5$. And hence the product is a multiple of the factorial.

Also, the factor is a binomial coefficient.

$\pi(R)=\displaystyle\frac{h!}{l!}$ and $\pi(R)=n!(X)$

\[\implies X=\frac{h!}{l!n!}=\frac{h!}{l!(h-l)!}={h\choose l}\]