# Quotient of factorials

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.

• This is the number of ways in which you can divide a group of $n^2$ objects into $n$ groups of $n$ objects each when there is no ordering within the groups or between the groups. – WimC Mar 24 at 6:41
• Thanks I understood – RUTAN GALENO Mar 24 at 6:44

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 $$(n,n,...,n).$$