Example 23 Find the polynomial $f(x)$ that satisfies $f\left(x^{n}+1\right)=f^{n}(x)+1$.

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$$
\begin{array}{l}
\text { Analysis: If } f(0)=0, \text { then } f(x)=x . \\
\quad \text { Let } f(x)=a_{m} x^{m}+a_{m-1} x^{m-1}+\cdots+a_{0}, a_{0} \neq \\
0, \varepsilon=e^{\frac{2 \pi i}{n}}, \\
\quad \text { then } f^{n}(\varepsilon x)=f\left((\varepsilon x)^{n}+1\right)-1=f\left(x^{n}+1\right) \\
-1=f^{n}(x),
\end{array}
$$

Analysis: If $f(0)=0$, then $f(x)=x$.
Let $f(x)=a_{m} x^{m}+a_{m-1} x^{m-1}+\cdots+a_{0}, a_{0} \neq$ $0, \varepsilon=e^{\frac{2 \pi i}{n}}$,

then $f^{n}(\varepsilon x)=f\left((\varepsilon x)^{n}+1\right)-1=f\left(x^{n}+1\right)$ $-1=f^{n}(x)$,

so there exists $0 \leqslant k<n$, such that $f(\varepsilon x)=\varepsilon^{k} f(x)$. Comparing the constant terms on both sides, we get $a_{0}=\varepsilon^{k} a_{0}$,
so $\varepsilon^{k}=1, f(\varepsilon x)=f(x)$.
Comparing the coefficients of each term, we get $a_{l} \varepsilon^{l}=a_{l}(1 \leqslant l \leqslant m)$, so when $n$ does not divide $l$, $a_{l}=0$. Thus, $f(x)=\varphi\left(x^{n}\right)$ $=\psi\left(x^{n}+1\right)$, where $\varphi, \psi$ are both polynomials.

Let $y=x^{n}+1$, then $\psi(y)=f^{n}(x)=f\left(x^{n}+1\right)$ $-1=f(y)-1=\psi\left(y^{n}+1\right)-1$, so $\psi(y)$ satisfies the same equation as $f(x)$, $\psi\left(y^{n}+1\right)=\psi^{n}(y)-1$, but the degree of $\psi$ is the degree of $f$ divided by $n$.

Continuing in this manner, if the constant term of each polynomial obtained is non-zero (i.e., $\psi(0) \neq 0, \cdots$), then the degrees of $f$, $\psi$, $\cdots$ are all multiples of $n$. Thus, $f$ can only be a constant (degree 0) $a_{0}$, and $a_{0}$ is a root of the equation $a_{0}=a_{0}^{n}+1$ (from which it is easy to see that $n$ must be an odd number greater than 1, and $a_{0}<-1$).

If $\psi(0)=0$, then $\psi(y)=y, f(x)=\psi\left(x^{n}+1\right)$ $=x^{n}+1$. Similarly, $f(x)=g(g(x))$, $g(g(g(x))), \cdots$, where $g(x)=x^{n}+1$.