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I am trying to figure out if the following expression

$$\frac{(n^2 - n)! }{ n! ((n-1)!)^n }$$

is an integer for all positive integer $n.$

I tried the induction, but induction case is running into problem. So I was looking at it from permutation/combination problems. But so far, couldn't come up with a convincing argument.

Does anybody has any thoughts on this to share ?

Thank you very much.

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closed as off topic by Todd Trimble, Anthony Quas, Dan Petersen, Emil Jeřábek, quid Jan 17 '13 at 17:43

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

This site is not the right place for your question; this site is for research interests of professional mathematicians. Try instead. – Todd Trimble Jan 17 '13 at 16:44

Todd was right, this is not appropriate for MO. Just an easy explanation: Let $\Sigma_n$ be the permutation group on $n$ letters. Then $\Sigma_m\wr\Sigma_n=(\Sigma_m)^n\rtimes \Sigma_n$ is a subgroup of $\Sigma_{mn}$. The above is a special case.

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More generally, $\frac{(nm)!}{n!(m!)^n}$ counts the partitions of a set of $nm$ elements into $n$ classes of $m$ elements.

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