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?

Note that $\dfrac{(a+b)!}{a! b!}$ is an integer. In general, $$\dfrac{\left(\displaystyle \sum_{k=1}^m a_k \right)!}{\displaystyle \prod_{k=1}^m a_k!}$$ is an integer. The above can be shown to be an integer by a combinatorial argument. Consider $\displaystyle\sum_{k=1}^m a_k$ balls, where $a_l$ balls are of color '$l$'. Then the number of possible arrangements of these balls in a straight line is given by $\dfrac{\left(\displaystyle \sum_{k=1}^m a_k \right)!}{\displaystyle \prod_{k=1}^m a_k!}$, which is an integer.

In your case, take $m=n$ and $a_k = n$, $\forall k \in \{1,2,\ldots,n\}$.

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 $n$.