How to prove this using combinatorics? How do you proceed if you are required to prove for any natural number $n$ that $$\frac{n^2!}{(n!)^n}$$ is an integer. Here the ! sign represents factorial. I got absolutely no leads on this problem. Any hint or help would be appreciated.
 A: How many ways are there to order a "word" made with $n$ letters, each of which occur $n$ times?
A: We'll show that the above number is the size of a set with integer number of elements.
Assume we have $n$ types of balls, and from each type we have $n$ balls.
Obviously, there are $n^2$ balls. Let's set them up in a line; obviously, there is an integer number of ways to do that.
First we set them all in a line as they are different- permutation over $n^2$ different elements is defined as $n^2!$, now divide by number of substitutions for each type - $n!$ for each type we get just what we wanted:
$$\frac{n^2!}{(n!)^n} $$
A: The product of $n$ consecutive integers is divisible by $n!$:
$$
(m+1)(m+2)\cdots(m+n)=\binom{m+n}{n}n!
$$
$n^2!$ is a product of $n$ products of $n$ consecutive integers:
$$
[1 \cdot 2 \cdots n]\dot[(n+1)(n+2)\cdots(n+n)]]\cdots
$$
Explicily,
$$
n^2! = \binom{n}{n}\binom{2n}{n}\cdots \binom{n^2}{n}(n!)^n
$$
A: a useful principle is that for any natural numbers $m,n$ we have:
$$
F_{n,0} | F_{n,m}
$$
where
$$
F(n,m) =  \prod_{j=1+m}^{n+m} j
$$
with this notation we have $F(n,0) = n!$ and
$$
(kn)! = \prod_{j=0}^{k-1} F(n,kn)
$$
setting $k=n$ gives the required result
