Does it follow that $(n!)^n$ divide $(n^2)!$

elementary-number-theory divisibility factorial

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.

Thanks,

-Larry


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

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