Prove that $${(n^2)!\over(n!)^{n+1}}$$ is an integer, where $n$ is a natural number greater than $5$.
I know how the product of $r$ consecutive numbers is divisible by $r!$ Could we use it here? If so how, if not please help with any other suitable method.
It's $$\frac{1}{n!}\binom{n^2}{n,n,...,n},$$ which is a natural number for all natural $n$ because there are $n!$ permutations exactly of $(n,n,...,n).$