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).

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
$$\binom{kn}{\underbrace{n,\cdots,n}_k}=\frac{(kn)!}{\underbrace{n!n!\cdots n!}_k}=\frac{(kn)!}{(n!)^k}$$
