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).

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.
