67th NMO Selection Tests for BMO and IMO

Let $n=k+lmk!$, where $l$ is an arbitrary nonnegative integer, let $p$ be any prime factor of $m$, and let $p^h$ be the highest power of $p$ that divides $k!$ — that is, $p^h$ divides $k!$ but $p^{h+1}$ does not. Notice that $n\equiv k(\mathrm{mod}{p}^{h+1})$, to deduce that $n(n-1)\cdots (n-k+1)\equiv k!(\mathrm{mod}{p}^{h+1})$, so $p^h$ is also the highest power of $p$ that divides the product $n(n-1)\cdots (n-k+1)$. Consequently, $p$ does not divide $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, so $m$ and $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ are indeed coprime.