1. Given an $n$-degree polynomial $p(x)$ satisfies $p(k)=\frac{1}{C_{n+1}^{k}}$, where $k=0,1,2, \cdots, n$. Find $p(n+1)$.
1. From $\Delta^{n+1} p(0)=0$, i.e., $\sum_{k=0}^{n+1}(-1)^{n+1-k} \cdot C_{n+1}^{k} \cdot p(k)=0$, we get $p(n+1)=\sum_{k=0}^{n}(-1)^{k}=$ $\left\{\begin{array}{l}1, \text { when } n \text { is even, } \\ 0, \text { when } n \text { is odd. }\end{array}\right.$