Solve $P_{n}$ in only one way, i.e., $\{1\},\{2\}, \cdots,\{n\}$. If we already have
$$
P_{n}, P_{n-1}, \cdots, P_{\left.k^{4}\right\}}
$$

each one is a refinement of the next, then we can combine any two sets in $P_{k+1}$ into one set to produce $P_{k}$, so $P_{k}$ has $C_{k-1}^{2}$ possibilities. Therefore, the number of partition chains of length $n$ is
$$
\prod_{k=1}^{n-1} C_{k+1}^{2}=\frac{(n-1)!n!}{2^{n-1}}
$$


Example 4 Find the number of partition chains of length $n$.

