Example 5 Let $f(x)$ be an $n$-degree polynomial, satisfying $f(k)=2^{k}, k=1,2, \cdots, n+1$. Try to find the values of $f(n+2)$, $f(n+3)$ (express the answers as complex functions of $n$ in their simplest form).
Analysis: Since the $(n+1)$-th order difference of $f(x)$ is zero, we get from formula (1) that $\sum_{k=0}^{n+1}(-1)^{n+1-k} C_{n+1}^{k} f(x+k)=0$. By taking $x=1$, 2, we can find $f(n+2), f(n+3)$.

Solution: Since $\operatorname{deg} f=n$, the $(n+1)$-th order difference of $f(x)$ is zero. Hence, from formula (1), we get $\sum_{k=0}^{n+1}(-1)^{n+1-k} C_{n+1}^{k} f(x+k)=0$. Taking $x=1$ and substituting $f(k)=2^{k} (k=1,2, \cdots, n+1)$, we obtain
$$
\begin{aligned}
f(n+2) & =-\sum_{k=0}^{n}(-1)^{n+1-k} C_{n+1}^{k} 2^{k+1}=2^{n+2}-2\left(\sum_{k=0}^{n+1}(-1)^{n+1-k} 2^{k} C_{n+1}^{k}\right) \\
& =2^{n+2}-2(2-1)^{n+1}=2^{n+2}-2 .
\end{aligned}
$$

Taking $x=2$, we can similarly find
$$
\begin{aligned}
f(n+3) & =-\sum_{k=0}^{n-1}(-1)^{n+1-k} C_{n+1}^{k} 2^{k+2}+C_{n+1}^{n}\left(2^{n+2}-2\right) \\
& =2^{n+3}-4 \sum_{k=0}^{n+1}(-1)^{n+1-k} 2^{k} C_{n+1}^{k}-2 C_{n+1}^{n} \\
& =2^{n+3}-4(2-1)^{n+1}-2(n+1) \\
& =2^{n+3}-2 n-6 .
\end{aligned}
$$