Show that $(n!)^{(n-1)!}$ divides $(n!)!$
$$ \begin{align} \frac{\displaystyle\left(\sum_{i=1}^na_i\right)!}{\displaystyle\prod_{i=1}^na_i!} &=\frac{\displaystyle\left(\sum_{i=1}^{n-1}a_i\right)!}{\displaystyle\prod_{i=1}^{n-1}a_i!} \frac{\displaystyle\left(\sum_{i=1}^na_i\right)!}{\displaystyle\left(\sum_{i=1}^{n-1}a_i\right)!\ a_n!}\\ &=\frac{\displaystyle\left(\sum_{i=1}^{n-1}a_i\right)!}{\displaystyle\prod_{i=1}^{n-1}a_i!} \binom{\displaystyle\sum_{i=1}^na_i}{a_n}\\ &=\prod_{k=1}^n\binom{\displaystyle\sum_{i=1}^ka_i}{a_k}\tag{1} \end{align} $$ Thus, the fraction on the left of $(1)$ is a product of binomial coefficients.
We can write $$ n!=n(n-1)!=\sum_{i=1}^{(n-1)!}n\tag{2} $$ Using $(2)$ and then $(1)$ $$ \begin{align} \frac{(n!)!}{n!^{(n-1)!}} &=\frac{\displaystyle\left(\sum_{i=1}^{(n-1)!}n\right)!}{\displaystyle\prod_{i=1}^{(n-1)!}n!}\\[4pt] &=\prod_{k=1}^{(n-1)!}\binom{kn}{n}\tag{3} \end{align} $$ Since the right hand side of $(3)$ is a product of binomial coefficients, the left hand side is an integer.
Example
The numbers get huge very quickly, but with $n=4$, $$ \frac{(4!)!}{4!^{3!}}=\frac{24!}{24^6}=3,246,670,537,110,000 $$ and $$ \begin{align} \prod_{k=1}^{3!}\binom{4k}{4} &=\binom{24}{4}\binom{20}{4}\binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}\\[4pt] &=3,246,670,537,110,000 \end{align} $$
[Three years later update] Addressing @Patrick’s comment, I should have said
Suppose you have $(n-1)!$ identical sets of $n$ balls. The balls in each set are numbered 1 through $n$. How many ...
[My original backwards answer follows.]
Suppose you have $n$ identical sets of $(n-1)!$ balls. The balls in each set are numbered 1 through $(n-1)!$. How many ways are there to arrange these $n!$ balls in a row? There are $\frac{(n!)!}{(n!)^{(n-1)!}}$ ways.
This is the same as showing that combinations or binomial coefficients $\displaystyle{n\choose k}=C_n^k=\prod_{j=0}^{k-1}\frac{n-j}{1+j}$ are always natural, for all values of n and k. It all depends on proving that the product of any k consecutive numbers is always divisible through the product of the first k consecutive numbers, $1$ through k. This is obvious, since, in each sequence of $2$ consecutive numbers, exactly one is even, and one odd; in each sequence of three consecutive numbers, exactly one is ternary, while the other two aren't; in each sequence of four consecutive numbers, exactly one is quaternary, while the rest aren't; etc. Our product, $(n!)!=1\cdot2\cdot3\cdot\ldots\cdot n!$, can be broken up into $\displaystyle\frac{n!}n=(n-1)!$ sequences of n consecutive terms, each such sub-product being divisible through n!, for the reasons explained above. In other words, the whole product is divisible through $(n!)^{(n-1)!}$. QED.