Show that $(n!)^{(n-1)!}$ divides $(n!)!$ 
Show that $(n!)^{(n-1)!}$ divides $(n!)!$

I found this question in a text he was reading about BREAKDOWN OF PRIME FACTORS IN FACT, I decided many years, have posted some here to help, but this ... Did not come out at all, in fact I could not leave the place, and looked reolhei theorems that shows me the text, but could not solve, need help.
 A: [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.
A: 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.
A: $$
\begin{align}
\frac{\displaystyle\left(\sum_{i=1}^na_i\right)!}{\displaystyle\prod_{i=1}^na_i!}
&=\frac{\displaystyle\left(\sum_{i=1}^{n-1}a_i\right)!}{\displaystyle\prod_{i=1}^{n-1}a_i!}
\frac{\displaystyle\left(\sum_{i=1}^na_i\right)!}{\displaystyle\left(\sum_{i=1}^{n-1}a_i\right)!\ a_n!}\\
&=\frac{\displaystyle\left(\sum_{i=1}^{n-1}a_i\right)!}{\displaystyle\prod_{i=1}^{n-1}a_i!}
\binom{\displaystyle\sum_{i=1}^na_i}{a_n}\\
&=\prod_{k=1}^n\binom{\displaystyle\sum_{i=1}^ka_i}{a_k}\tag{1}
\end{align}
$$
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
$$
\begin{align}
\prod_{k=1}^{3!}\binom{4k}{4}
&=\binom{24}{4}\binom{20}{4}\binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}\\[4pt]
&=3,246,670,537,110,000
\end{align}
$$
