I'd like to see if I understand what's discussed in the thread below:
prove that $(n^2)!$ is divisible by $(n!)^{n+1}$, where $n\in \mathbb{N}$
Here's what I think:
Let $n^2, n^2 - 1, n^2 - 2, \ldots, 1$ stand for balls. We can distribute the balls into $n+1$ distinct boxes $n!$ at a time. Then the number of such distributions is $\binom{(n^2)!}{n!, n!,\ldots,n!}.$ This overcounts the number of partitions of $\{n^2, n^2 - 1, n^2 - 2, \ldots, 1\}$ into $n+1$ sets $n!$ at a time by a factor of $(n+1)!$ Thus, $\frac{1}{(n+1)!}\binom{(n^2)!}{n!, n!,\ldots,n!}$ is the number of partitions of $\{n^2, n^2 - 1, n^2 - 2, \ldots, 1\}$ into $n+1$ sets $n!$ at a time. Since $\frac{1}{(n+1)!}\binom{(n^2)!}{n!, n!,\ldots,n!}$ is an integer, $(n^2)!$ is divisible by $(n!)^{n+1}$.
Does that make sense?
I think I get it: You want to distribute $n^2$ balls into $n$ boxes, putting $n$ of them in each box. Keeping the boxes distinct, we can count this as $\binom{n^2}{n,\ldots,n}=\frac{n^2!}{n!\cdots n!}$. Now, if the boxes are indistinguishable, we realize we've overcounted partitions into $n$ sets of $n$ by a factor of $n!$, so we can divide by $n!$ and still obtain an integer. Thus $n^2!$ is divisible by $n!\cdot(n!)^n = (n!)^{n+1}$. That makes sense to me.
I think you had the idea, but your notation was off. The multinomial coefficient is written without the factorials in it, and be careful about how many boxes there are, and how many balls of the $n^2$ we can put in each box in a particular partition.