Example 4 Let the $n$-degree polynomial $f(x)$ satisfy $f(k)=\frac{1}{C_{n+1}^{k}}, k=0,1,2, \cdots, n$, find $f(n+1)$.


Solution 1: Since $f(x)$ is known at $n+1$ distinct points, by the Lagrange interpolation formula, we can uniquely determine
$$
\begin{array}{l}
f(x)=\sum_{k=0}^{n} \frac{1}{C_{n+1}^{k}} \prod_{\substack{i \neq k \\
1 \leqslant i \leqslant n}} \frac{x-i}{k-i}=\sum_{k=0}^{n} \frac{\prod_{\substack{i \neq k \\
1 \leqslant i \leqslant n}}(x-i)}{C_{n+1}^{k}(-1)^{n-k}(n-k)!k!} \\
=\sum_{k=0}^{n}(-1)^{n-k} \frac{n+1-k}{(n+1)!} \prod_{\substack{i \neq k \\
1 \leqslant i \leqslant n}}(x-i), \\
\end{array}
$$

Therefore, $f(n+1)=\sum_{k=0}^{n}(-1)^{n-k} \frac{n+1-k}{(n+1)!} \prod_{\substack{i \neq k \\ 1 \leqslant i \leqslant n}}(n+1-i)$
$$
\begin{array}{l}
=\sum_{k=0}^{n} \frac{(-1)^{n-k}}{(n+1)!} \prod_{i=0}^{n}(n+1-i)=\sum_{k=0}^{n}(-1)^{n-k} \\
=\left\{\begin{array}{l}
0, \text { when } n \text { is odd } \\
1, \text { when } n \text { is even }
\end{array}\right.
\end{array}
$$

Solution 2: Since $\operatorname{deg} f=n$, the $(n+1)$-th difference of $f(x)$ is 0, so by property (iii), we get $\sum_{i=0}^{n+1}(-1)^{i} C_{n+1}^{i} f(i)=0$. Substituting $f(i)=\frac{1}{C_{n+1}^{i}}(i=0,1, \cdots, n)$, we obtain $f(n+1)=$ $\sum_{i=0}^{n}(-1)^{n-i}$.
This yields the result from Solution 1.