$5 \cdot 7$ Let $n$-degree polynomial $p(x)$ satisfy $p(k)=\frac{1}{C_{n}^{k}}$, where $k=0,1,2, \cdots$, $n$. Find $p(n+1)$.
［Solution］ According to the Lagrange interpolation formula, we get
$$p(x)=\sum_{k=0}^{\prime \prime} \frac{1}{C_{n+1}^{k}} \prod_{\substack{i \neq k \\ 0 \leqslant i \leqslant n}} \frac{x-i}{k-i}$$

But $\prod_{\substack{i \neq k \\ 0 \leqslant \leqslant}}(k-i)=k(k-1) \cdots[k-(k-1)] \cdot[k-(k+1)] \cdots(k-n)$
Therefore,
$$\begin{array}{l}
=\sum_{k=0}^{n}(-1)^{n-k} \frac{n+1-k}{(n+1)!} \prod_{\substack{i \neq k \\
0 \leq \text{heart} \leq}}(x-i), \\
p(n+1)=\sum_{k=0}^{n}(-1)^{n-k} \frac{n+1-k}{(n+1)!} \prod_{\substack{i \neq k \\
0 \leq \leqslant}}(n+1-i) \\
=\sum_{k=0}^{n} \frac{(-1)^{n-k}}{(n+1)!} \prod_{i=0}^{n}(n+1-i) \\
=\sum_{k=0}^{n}(-1)^{n-k} .
\end{array}$$

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

Note: The symbol "heart" in the original text seems to be a typographical error and should be "i".