I need to prove the following equation.
$$\frac{(n!)!}{(n!)^{(n-1)!}} = x , x\in \mathbb Z$$
In other words, I want to prove that $\frac{(n!)!}{(n!)^{(n-1)!}}$ is an integer. It can be solved easily using Mathematical Induction, but I want to prove it by other methods.
Combinatorial proof:
Consider the number of ways to divide $n!$ objects in $(n-1)!$ groups of $n$ objects each, where the ordering in each group does not matter. This is clearly an integer.
Now notice that the number of ways to do this where the order does matter in each group is $(n!)!$, and the number ways to order objects in each group is $n!$. Since the number of groups is $(n-1)!$, the number of ways to divide $n!$ objects in $(n-1)!$ groups of $n$ objects each where the ordering does not matter is equal to $\frac{(n!)!}{(n!)^{ (n-1)! }}$, which is hence an integer.
HINT: Apply Legendre's theorem and note that $\lfloor nx\rfloor\geq n\lfloor x\rfloor$ for positive $n$ and $x$.
Consider any prime factor $p$, we will show that $p$ divides the numerator more times than it divides the denominator.
The number of times $p$ divides $k!$, for any $k$, is given by: $$ \sum_{i \ge 1} \left\lfloor \frac{k}{p^i} \right\rfloor = \left\lfloor \frac{k}{p} \right\rfloor + \left\lfloor \frac{k}{p^2} \right\rfloor + \left\lfloor \frac{k}{p^3} \right\rfloor + \left\lfloor \frac{k}{p^4} \right\rfloor + \cdots. $$ Therefore, the number of factors of $p$ in the numerator is $$ \sum_{i \ge 1} \left\lfloor \frac{n!}{p^i} \right\rfloor \tag{1} $$ And the number of factors of $p$ in the denominator is $$ (n-1)! \cdot \sum_{i \ge 1} \left\lfloor \frac{n}{p^i} \right\rfloor = \sum_{i \ge 1} (n-1)! \left\lfloor \frac{n}{p^i} \right\rfloor $$ The result follows term-by-term, since $$ (n-1)! \left\lfloor \frac{n}{p^i} \right\rfloor \le \left\lfloor \frac{(n-1)! \cdot n}{p^i} \right\rfloor = \left\lfloor \frac{n!}{p^i} \right\rfloor . $$
