Conjecture for product of binomial coefficient Is it true that for any $n, k\in\mathbb N$
$$\frac{(kn)!}{k!(n!)^k} = \prod_{l=1}^k {{ln-1}\choose{n-1}} \quad?$$
I tested it for some small $k$ and $n$, but I don't know how to prove that it is true (or find example showing that it is not).
 A: We have 
$${ln -1 \choose n-1} = \frac{(ln-1)!}{(n-1)! [(l-1)n]!} = \frac{1}{n!}\times \frac{(ln-1)!}{ [(l-1)n - 1]!} \times \frac{1}{l-1}$$
so
$$\prod_{l=1}^k {ln -1 \choose n-1} = \frac{n}{n!^k}\times \prod_{l=2}^k\frac{(ln-1)!}{[(l-1)n - 1]!} \times \prod_{l=2}^k\frac{1}{l-1}$$
In the middle product successive terms cancel
$$\prod_{l=2}^k\frac{(ln-1)!}{[(l-1)n - 1]!}  = \frac{(kn-1)!}{(n-1)!}$$
and
$$\prod_{l=2}^k\frac{1}{l-1} = \frac{1}{(k-1)!}$$
so
$$\prod_{l=1}^k{ln -1 \choose n-1} =  \frac{n}{n!^k} \frac{(kn-1)!}{(k-1)!} =  \frac{(kn)!}{(n!)^kk!}$$
proving the claim.
A: $\textbf{1)}$ If $k=1$, we have $\frac{n!}{n!}=\binom{n-1}{n-1}=1$.
$\textbf{2)}$ Suppose $\displaystyle\frac{(kn)!}{k!(n!)^k} = \prod_{l=1}^k {{ln-1}\choose{n-1}} \quad$ for some $k\ge1$.
Then $\displaystyle\frac{((k+1)n)!}{(k+1)!(n!)^{k+1}}=\left(\frac{kn+n}{(k+1)n}\right)\frac{(kn+n-1)!}{(n-1)!}\frac{1}{k!(n!)^{k}}=\frac{(kn+n-1)!}{(n-1)!(kn)!}\frac{(kn)!}{k!(n!)^{k}}$
$\displaystyle=\binom{(k+1)n-1}{n-1}\prod_{l=1}^{k}\binom{ln-1}{n-1}=\prod_{l=1}^{k+1}\binom{ln-1}{n-1}.$
A: To show that the theorem is true by algebraic manipulations, observe that
\begin{align}
\frac{(kn)!}{(n!)^k}
=\frac{(kn)!}{(kn-n)!n!}\cdot\frac{(kn-n)!}{(n!)^{k-1}}
&=\binom{kn}{n} \frac{(kn-n)!}{(n!)^{k-1}}\\
&=\binom{kn}{n}\binom{kn-n}{n}\cdot\frac{(kn-2n)!}{(n!)^{k-2}}\\
&\;\vdots\\
&=\binom{kn}{n}\binom{kn-n}{n}\cdots \binom{kn-mn}{n}\cdot \frac{(kn-(m+1)n)!}{(n!)^{k-(m+1)}}\\
&\;\vdots\\
&=\binom{kn}{n}\binom{kn-n}{n}\cdots \binom{2n}{n}\binom{n}{n}
\end{align}
Note that the last fraction has been exhausted in this manner. Thus the LHS of the identity may be rewritten as 
$$\frac{(kn)!}{k! (n!)^k}=\frac{1}{k!}\prod_{l=1}^{k}\binom{ln}{n}=\prod_{l=1}^{k}\frac{1}{l}\frac{(ln)!}{n! (ln-n)!}\cdot\frac{n}{n}=\prod_{l=1}^{k}\binom{ln-1}{n-1}$$
as claimed.
One can also do a counting proof of the first identity I showed above: Suppose we want to form $kn$ people into $n$ teams of $k$. One way to do that is first choose $n$ out of $kn$ people, then the next $n$, etc. until all $k$ teams are selected. This can be done in $\binom{kn}{n}\cdots \binom{kn-n}{n}\cdots\binom{n}{n}=\prod_{l=1}^k \binom{ln}{n}$ ways. But I could also do this all at once, in which case the total number is given by the multinomial coefficient
$$\binom{kn}{\underbrace{n,\cdots,n}_k}=\frac{(kn)!}{\underbrace{n!n!\cdots n!}_k}=\frac{(kn)!}{(n!)^k}$$
A: Using the fact that 
$$\frac{\bigg((n-1)!\bigg)^k}{(n!)^k} = \frac{1}{n^k},$$
the formula becomes:
$$\frac{(kn)!}{k!n^k} = \prod_{l = 1}^k \frac{(ln-1)!}{(ln-n)!} = \prod_{l=1}^k \prod_{j=1}^{n-1} (ln-j).$$
By simply looking at the forumla on the right, you will see that it is precisely the product $(kn) \cdot (kn-1) \cdot (kn-2) \cdot \dots \cdot (1)$ except that it is missing the terms of the form $ln$ for $l = 1$ to $k$. The product of these missing terms is $k!n^k$, and so the formula is valid.
