Prove that $\dfrac{(n^2)!}{(n!)^n}$ is an integer for every $n \in \mathbb{N}$ Prove that $\dfrac{(n^2)!}{(n!)^n}$ is an integer for every $n \in \mathbb{N}$ 
I know that there are tools in Number theory to proves the required but I want to use the tool that says that if you can prove that an expression solves a combinatorial problem then it represents an integer for every $n \in \mathbb{N}$
My solution is, assume we are given $n^2$ beads, $n$ beads in every color of $n$ colors, and we want to place them in a row ($(n^2)!$)
now since we have $n$ groups of colors and every group has $n$ beads in it, we need to cancel the inner sort of each of the groups, which gives us $(n!)^n$ for all the possible inner sorts of the beads.
Will that proof work? What are its flaws?

The argument can be continued to give an additional $n!$ of divisibility.
After removing the redundancy of the inner ordering within the groups, the groups can be sorted relative to each other (by least element, or in any other way that gives a unique order).  
Hence $\frac{(n^2)!}{(n!)^{n+1}}$ is an integer, equal to the number of partitions of $n^2$ different objects into $n$ sets of $n$.
