Example 3 Let $P_{m}=\left\{A_{1}, A_{2}, \cdots, A_{m}\right\}$ be a partition of an $n$-element set $X$ (i.e., $A_{1}, A_{2}, \cdots, A_{m}$ is a partition of $X$). If one of the $A_{i}$ is further divided into two sets, this results in a partition $P_{m+1}$ of $X$, which consists of $m+1$ sets. $P_{m+1}$ is called a refinement of $P_{m}$. If
$$
P_{1}, P_{2}, \cdots, P_{n}
$$

are all partitions of the $n$-element set $X$, and each is a refinement of the previous one (clearly, in this case, $P_{n}$ consists of $n$ sets, while $P_{1}$ consists of only one set, which is $X$), then (7) is called a chain of length $n$. Find the number of chains of length $n$.
Starting from $X$, we can generate a chain of length $n$ (7) by gradually refining it. This process can also be reversed: starting from a partition $P_{n}$ consisting of $n$ singleton sets, we can merge two of these sets to obtain $P_{n-1}$, then merge two sets in $P_{n-1}$ to get $P_{n-2}$, and so on. Generally, suppose we already have $P_{n}, P_{n-1}, \cdots, P_{k+1}$. By merging any two sets in $P_{k+1}$, we obtain $P_{k}$. Since $P_{k+1}$ consists of $k+1$ sets, there are $C_{k+1}^{2}$ ways to form $P_{k}$. Therefore, the number of chains of length $n$ is
$$
\prod_{k=1}^{n-1} C_{k+1}^{2}=\frac{n!(n-1)!}{2^{n-1}}
$$

types.