1. There are $n$ points $(n>1)$, sequentially denoted as $P_{1}, P_{2}, \cdots, P_{n}$. Connecting these $n$ points, find the number of ways to connect them such that the polyline $P_{1} P_{2} \cdots P_{n}$ does not intersect itself.
Let's first consider the situation where the position of point $P_{n}$ is determined. In this case, there are $2^{n-2}$ ways to connect the points.

When $n=2$, if we fix one point as $P_{2}$, the other point must be $P_{1}$, giving 1 way (when $n=2$, $2^{n-2}=1$). Now, we need to connect $P_{1}$ and $P_{n}$: either by a straight line or by the path “$P_{1} P_{2} \cdots P_{n} P_{n+1}$”, which is the same as “$P_{n+1} P_{1} P_{2} \cdots P_{n}$”, giving 2 possible ways to connect. This leads to: for $n+1$ points, there are $2 \cdot 2^{n-2}$ ways to connect $\left(2 \cdot 2^{n-2}=2^{(n+1)-2}\right)$. In this situation, there are $2^{n-2}$ ways to connect. Since the position of $P_{n}$ has $n$ possible choices, the number of ways to connect the polyline $P_{1} P_{2} \cdots P_{n}$ that satisfies the problem's requirements is $n \cdot 2^{n-2}$.