How to prove that $\frac{(12!)!}{12!^{11!}}$ is integer? So far I have used that a combination is an integer so $\frac{n!}{m!(n-m)!}$ is integer. Now let $n=mb$ so $\frac{mb!}{m!(mb-m)!}$. 
What is left is to prove that $\frac{(mb)!}{m!^b}$ is integer so that i can apply $m=12$ and $b =11!$
How to do that?
 A: $\frac{(mb)!}{m!^b}$ counts the number of ways of putting $mb$ labeled balls into $b$ labeled boxes, with $m$ balls in each box.
To see this, note that any of the $(mb)!$ ways of ordering the balls gives us a way to put the balls in the boxes, but that each configuration arises from $m!^b$ possible orderings (all the ways of rearranging the balls within the boxes).
A: A slightly different way to express JimmyK4542’s answer is to notice that $\frac{(mb)!}{m!^b}$ is a multinomial coefficient:
$$\frac{(mb)!}{m!^b}=\binom{mb}{\underbrace{m,\,m,\,\ldots,\,m}_{b\text{ copies}}}\;,$$
which is the coefficient of $x_1^mx_2^m\ldots x_b^m$ in $(x_1+x_2+\ldots+x_b)^{mb}$; this coefficient is of course a non-negative integer.
Equivalently, this multinomial coefficient can be interpreted as the number of ways to partition the set $[mb]=\{1,2,\ldots,mb]$ into $b$ labelled sets $A_1,\ldots,A_b$, each of size $m$, just as $\binom{2n}n$ is the number of ways to partition $[2n]$ into labelled sets $A_1$ and $A_2$ (by counting the ways to choose $A_1$). First, there are $(mb)!$ ways to list the elements of $[mb]$, and we can then put the first $m$ elements of the list into $A_1$, the second $m$ into $A_2$, and so on. However, for a given partition $\{A_1,\ldots,A_b\}$ the elements of each $A_k$ could have been listed in any of $m!$ different orders, so there are $m!^b$ different permutations of $[mb]$ that yield the partition $\{A_1,\ldots,A_b\}$. Thus, the number of distinct partitions is $\frac{(mb)!}{m!^b}$. This is essentially Slade’s solution.
A: We know that product of $n$ consecutive integer is divisible by $n!$
So Here  $1.2.3.4........................................n$ is divisible by $n!$
Similarly $(n+1)\cdot(n+2)\cdot(n+3)...........(2n)$ is divisible by $n!$
Similarly $(2n+1)\cdot (2n+2)\cdot ..............(3n)$ is divisible by $n!$
Similarly $(3n+1)\cdot(3n+2)......................(4n)$ is divisible by $n!$
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Similarly $\{(k-1)n+1\}\cdot \{(k-1)n+2\}\cdot..................\{(k-1)n+n\}$ is divisible by $n!$
So we can say that $(kn)!$ is divisible by $(n!)^k$
Now Put $k=(n-1)!\;,$ We get $(n!)!$ is divisible by $(n!)^{(n-1)!}$
A: You have the right idea, you just need to take it a bit further. Notice that 
$\dfrac{(mb)!}{m!^b} = \dfrac{(mb)!}{m![m(b-1)]!} \cdot \dfrac{[m(b-1)]!}{m![m(b-2)]!} \cdot \dfrac{[m(b-2)]!}{m![m(b-3)]!} \cdots \dfrac{[3m]!}{m![2m]!} \cdot \dfrac{[2m]!}{m![m]!}$
$= \dbinom{mb}{m} \cdot \dbinom{m(b-1)}{m} \cdot \dbinom{m(b-2)}{m} \cdots \dbinom{3m}{m} \cdot \dbinom{2m}{m}$.
Do you see why this must be an integer?
A: In fact, if you know a little group theory, you can do a little better. As other have remarked, it is always true that $(a!)^{b}$ divides $(ab)!$ for positive integers $a$ and $b$. However, it is even true that $b! (a!)^{b}$ divides $(ab)!$.
(Corrected version): This can be see using Lagrange's Theorem for the symmetric group $S_{ab}$. This symmetric group has a subgroup $S_{a} \wr S_{b}$, which is a semidirect product of a "base group" which is itself a direct product of $b$ copies of $S_{a}$, and a subgroup isomorphic $S_{b}$ which permutes these direct factors around in the natural manner. This subgroup has order $(a!)^{b}
By Lagrange's Theorem, $b!(a!)^{b}$ must be a divisor of $(ab)!$.
Applying this with $a = 12$ and $b = 11!$ gives that we even have that 
$(11!)! (12!)^{11!}$ divides $(12!)!$.
