Example 6 There are $n(>1)$ points on a circle, connect these $n$ points, sequentially denoted as $p_{1}, p_{2}, \cdots, p_{n}$, such that the polyline $p_{1} p_{2} \cdots p_{n}$ does not intersect itself. How many such connection methods are there?
(1986 Former Federal Republic of Germany Competition Problem)
When $n=2$, connecting $p_{1} p_{2}$ has only 1 method.
When $n=3$, if we fix $p_{3}$ as the endpoint, then $p_{3}$ can only connect to $p_{1}$ or $p_{2}$, so the number of connection methods with $p_{3}$ as the endpoint is $1 \cdot 2=2=2^{3-2}$. Suppose the number of connection methods with $p_{n}$ as the endpoint is $2^{n-2}$. By symmetry, the number of connection methods with $p_{1}$ as the endpoint is also $2^{n-2}$. When adding another point $p_{n+1}$ as the endpoint, $p_{n+1}$ can only connect to $p_{1}$ or $p_{n}$, so the number of connection methods with $p_{n+1}$ as the endpoint is $2^{n-2}+2^{n-2}=2^{n-1}$. This proves by mathematical induction that the number of connection methods with $p_{n}$ as the endpoint is $2^{n-2}$. Since each of the $n$ points can be the endpoint, the total number of methods (including repeated counts) is $n \cdot 2^{n-2}$. When $n \geqslant 3$, each connection method is counted twice, so the number of valid methods is $\left\{\begin{array}{l}1(n=2 \text { when }), \\ n \cdot 2^{n-3}(n \geqslant 3 \text { when). }\end{array}\right.$