Example 2 If the partition $A_{1}, A_{2}, \cdots, A_{m}$ of an $n$-element set $X$ contains $k_{1}$ singletons, $k_{2}$ pairs, $\cdots \cdots, k_{n}$ $n$-element sets $\left(k_{1}+k_{2}+\cdots+k_{n}=m\right.$, $1 k_{1}+2 k_{2}+\cdots+n k_{n}=n, k_{1}, k_{2}, \cdots, k_{n}$ are non-negative integers $)$, then this partition is called a partition of the form $1^{k_{1}} \cdot 2^{k_{2}} \cdots \cdots \cdot n^{k_{n}}$. Find the number of such partitions.
For each partition of the form $1^{k_{1}} \cdot 2^{k_{2}} \cdots \cdots \cdot n^{k_{n}}$, the elements can be arranged as follows:

First, arrange the elements of the one-element sets (there are $k_{1}$! ways to do this), then arrange the elements of the two-element sets, with the order of each set being $k_{2}$! and each set having 2! ways to arrange its elements, resulting in a total of $(2!)^{k_{2}} \cdot k_{2}!$ arrangements. Proceeding similarly, $k_{j}$ $j$-element sets have $(j!)^{k_{j}} \cdot k_{j}!$ arrangements.

Each different partition of the form $1^{k_{1}} \cdot 2^{k_{2}} \cdot \cdots \cdot n^{k_{n}}$ has at least one different set, so the arrangements produced by the above method are all distinct.

On the other hand, for any permutation of $n$ elements, the first $k_{1}$ elements form $k_{1}$ one-element sets, the next $2 k_{2}$ elements form $k_{2}$ two-element sets (each pair of consecutive elements forms one set), and so on, resulting in a partition of the form $1^{k_{1}} \cdot 2^{k_{2}} \cdots \cdots \cdot n^{k_{\pi}}$. The given permutation is precisely the arrangement produced by this partition using the above method.

Thus, the above method produces exactly all $n$! permutations, with no repetitions and no omissions. Therefore, the number of partitions of the form $1^{k_{1}} \cdot 2^{k_{2}} \cdot \cdots \cdot n^{k_{n}}$ is