Number of permutations with given cyclic structure If $\sigma$ is a permutation made up by the disjoint cycles $\tau_1, \dots, \tau_r$ (including those of length $1$), we call structure of $\sigma$ $$(l_1, \dots, l_r),$$ where $l_1, \dots, l_r$ are the lengths of $\tau_1, \dots, \tau_r$ (supposing we ordered them in such a way that $l_1 \ge l_2 \ge \dots \ge l_r$).

The number of permutations $\sigma \in S_n$ with structure $$(\underbrace{l_1, \dots, l_1}_{k_i \text{ times}}, \dots, \underbrace{l_m, \dots, l_m}_{k_m \text{ times}} ),$$ where $l_1 > \dots > l_m \ge 1$ and $h_1l_1 + \dots + h_ml_m = n$, is
$$\frac{n!}{\prod_{i=1}^m { {k_i}! {l_i}^{k_i}}}.$$

How does one prove this statement? I'm only aware of the proof that there are $$\frac{1}{l}\frac{n!}{(n-l)!}$$ $l$-cycles in $S_n$.

Suppose you are given a blank structure
$$
( \cdot \cdot \dots \cdot) \cdots ( \cdot \cdot \dots \cdot)
$$
of the given shape (partition).
Start filling the blanks in all possible ways. There are clearly $n!$ ways of doing this.
But of course you have overcounted.
First, a cycle of length $l$ can be written in $l$ different ways
$$
(1 2 3 \dots l) = (2 3 \dots l 1) = (3 4 \dots l 1 2) = \dots = (l 1 2 \dots l-1).
$$
You can do this for every cycle, so this accounts for the factor 
$$
\prod_{i=1}^{m} { {l_i}^{k_i}}
$$
in the denominator.
And then, the $k_{i}$ cycles of length $l_{i}$ can be permuted in all the possible $k_{i} !$ ways without changing the permutation, as they are disjoint and thus commute. This accounts for the factor
$$
\prod_{i=1}^m { {k_i}! }
$$
in the denominator.
