It is well known that $(n!)^2$ divides $(2n)!$.

Does it follow that $(n!)^3$ divides $(3n)!$ and so on up to $(n!)^n$ dividing $(n^2)!$?

If yes or no, could you provide the details behind the argument?

As a quick test, I tried $n=3$ and $n=4$ which worked.



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    $\begingroup$ You conjecture holds for every $n$ up to $200$. So I think it's reasonable to expect that it is true. $\endgroup$ – rubik Feb 4 '15 at 21:46

Yes, it's true. The simple argument: $\displaystyle \frac{(kn)!}{(n!)^{k}}$ for $1\leq k \leq n$ is equal multinomial coefficient: $$\displaystyle {kn \choose n,n,\cdots,n}$$.

  • $\begingroup$ Thanks very much. I had never heard about multinomials before. Fascinating. $\endgroup$ – Larry Freeman Feb 4 '15 at 21:51

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