This is straightforward. We have
$\displaystyle \binom{n}{k}\binom{k}{l}=\frac{n!}{k!(n-k)!}\cdot \frac{k!}{l!(k-l)!}
$ and $\displaystyle
\binom{n}{l}\binom{n-l}{k-l}=\frac{n!}{l!(n-l)!}\cdot \frac{(n-l)!}{(k-l)!(n-l-(k-l))!}
$
If you simplify these two expressions you'll see that they are equal.