(n²)! is divisible by (n!)^(n+1)

Idea. Interpret the quotient as the number of something, which is automatically an integer.

Counting. Let $N$ be the number of ways to partition a set of $n^2$ distinct objects into $n$ unlabelled groups of size $n$. Line the $n^2$ objects up in a row in any of $(n^2)!$ orders, and cut the row into consecutive blocks of $n$: this gives a partition. Each partition arises from exactly $(n!)^n\cdot n!$ orderings: $n!$ orders inside each of the $n$ groups, and $n!$ orders in which the groups appear. Hence
$$N=\frac{(n^2)!}{(n!)^n\cdot n!}=\frac{(n^2)!}{(n!)^{n+1}}.$$

Since $N$ is a number of objects, it is a positive integer.

Check. $n=2$: $\frac{4!}{(2!)^3}=\frac{24}8=3$, the three ways to split four people into two pairs.

Remark. More generally $\dfrac{(kn)!}{(n!)^kk!}$ is an integer.

Lesson. To prove that an expression is an integer, find a set of objects that it counts.