Suppose there exists prime $q$ that $\nu_q(n)>\nu_q(k!)$. Note that $q\leqslant n$. We have $(n+1)(n+2)\cdots (n+k)\equiv k!\pmod{q^{\nu_q(k!)+1}}\implies q\mid p\implies p=q$. So, $$\frac{(n+1)(n+2)\cdots (n+k)}{k!}-1\leqslant n\implies \frac{(k+1)!}{2}\leqslant (n+2)(n+3)\cdots (n+k)\leqslant k!\implies k\leqslant 1.\quad \perp$$