Example 5 Let the $n$-degree polynomial $P(x)$ satisfy $P(k)=\frac{1}{C_{n}^{k}}$, where $k=0,1,2, \cdots, n$, find $P(n+1)$.


$$
\begin{aligned}
P(x) & =\sum_{k=0}^{n} \frac{1}{C_{n+1}^{k}} \prod_{\substack{i \neq k \\
1 \leqslant i \leqslant n}} \frac{x-i}{k-i}=\sum_{k=0}^{n} \frac{\prod_{i \neq k}(x-i)}{C_{n+1}^{k}(-1)^{n-k}(n-k)!k!} \\
& =\sum_{k=0}^{n}(-1)^{n-k} \frac{(n+1-k)}{(n+1)!} \prod_{\substack{i \neq k \\
1 \leqslant i \leqslant n}}(x-i), \\
\therefore \quad P(n+1) & =\sum_{k=0}^{n}(-1)^{n-k} \cdot \frac{(n+1-k)}{(n+1)!} \prod_{\substack{i \neq k \\
1 \leqslant \leqslant \leqslant}}(n+1-i) \\
& =\sum_{k=0}^{n} \frac{(-1)^{n-k}}{(n+1)!} \prod_{i=0}^{n}(n+1-i)=\sum_{k=0}^{n}(-1)^{n-k} .
\end{aligned}
$$

Therefore, when $n$ is odd, $P(n+1)=0$, and when $n$ is even, $P(n+1)=1$.