Why $r!$ divides the integer $\frac{(rn)!}{(n!)^r}$ and why $\frac{(rn)!}{(n!)^r}$ is integer. During my reading for some lecture notes, I found the following statement :
It is clear that $r!$ divides the integer $\dfrac{(rn)!}{(n!)^r}$ , where $r$ and $n$ are positive integers.
I am not sure why $\dfrac{(rn)!}{(n!)^r}$ is integer and why it is divisible by $r!$.
I tried to figure it out by numbers but I could not see how to prove it.
Thanks 
 A: How many ways are there to distribute distinguishable $rn$ objects into $r$ labeled bins, so that there are $n$ objects in each bin?  There are $(rn)!$ ways to arrange the objects in a line, and then we put the first $n$ object in the first bin, the next $n$ objects in the second bin, and so on.  But we don't care in what order the objects are placed in each bin, so we have to divide by $(n!)^r$.  Thus, $(rn)!/(n!)^r$ is an integer.  Now suppose we don't distinguish between the bins.  Then to count the number of ways, we have to divide the preceding number by $r!$.   
A: You can prove by double-induction that $$\dfrac{(rn)!}{(n!)^r} = \binom{rn}{n}\times \binom{rn-n}{n}\times\binom{rn-2n}{n}\times\cdots\times\binom{rn-kn}{n}\times\dots\times \binom{n}{n} = \prod\limits_{k=0}^{r-1}\binom{rn-kn}{n}$$
We know each of these terms on the right to be integers and so as $\dfrac{(rn)!}{(n!)^r}$ must be an integer too as it is a product of integers.
As for $r!$ dividing this., recognize that this expression counts something... namely the number of ways to distribute $rn$ people into $r$ labeled groups with $n$ people in each group.  You can recognize then that as there are $r!$ different ways to rearrange the labels of the groups, the total number of arrangements must be a multiple of $r!$.
