Let $p|n+1$ is prime divisor of $n+1$ and $p \leq n$ $\frac{n!}{p!(n-p+1)!}=\frac{(n-p+2)(n-p+3)...n}{p!}$ so $p| (n-p+2)...n$ but $(n-p+2)...n \equiv 1*2*...(p-1) \neq 0 \pmod {p}$ So enough to check $n=p-1$ for prime $p$ $\frac{n!}{i!(n-i+1)!}=\frac{1}{n+1}\binom{n+1}{i}=\frac{1}{p} \binom{p}{i}$ is integer, because $p|\binom{p}{i}$