$1 \cdot 71$ A circle has $n$ points $(n>1)$, 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?
[Solution] We first use mathematical induction to prove: Given $P_{n}$ points, there are $2^{n-2}$ possible ways to connect them.
When $n=2$, then connecting $P_{1} P_{2}$ has only one method, which is $1=2^{2-2}$ ways.
Assume for $n$ points there are $2^{n-2}$ ways to connect, and then add one more point $P_{n+1}$.
Since for each way to connect $P_{n+1}$, and $P_{n+1}$ must be between $P_{1}$ and $P_{n}$, the formed polyline is either $P_{1} P_{2} \cdots P_{n} P_{n+1}$, or $P_{n+1} P_{1} P_{2} \cdots P_{n}$, i.e., there are 2 possible choices.

From this, we can conclude that for $n+1$ points, there are $2 \cdot 2^{n-2}=2^{(n+1)-2}$ ways to connect, so the proposition holds for $n+1$.
From the above, for $n>1$ points, when $P_{n}$ is determined, there are $2^{n-2}$ ways to connect.
Since there are $n$ possible choices for $P_{n}$, the number of connection methods that meet the requirements of the problem is $n \cdot 2^{n-2}$.