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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.

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    $\begingroup$ 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. $\endgroup$ – WimC Mar 24 at 6:41
  • $\begingroup$ Thanks I understood $\endgroup$ – RUTAN GALENO Mar 24 at 6:44
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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 $(n,n,...,n).$

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  • $\begingroup$ Ok to avoid the permutations of each group we are dividing by n! to obtain only the selections here divisions. Ok thanks I understand. $\endgroup$ – RUTAN GALENO Mar 24 at 6:52
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Mar 24 at 6:53

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