Example 3 Given $f(x)=\frac{x}{a+b x}$, find $f^{(n)}(x)$
$$
\begin{array}{l}
f^{11}(x)=f(x)=\frac{x}{a+b x} \cdot f^{2 \prime}(x)=f(f(x))=\frac{\frac{x}{a+b x}}{a+b \frac{x}{a+b x}}=\frac{x}{a^{2}+b x(1+a)} \\
f^{(31}(x)=\frac{x}{a^{3}+b x\left(1+a+a^{2}\right)} \cdots
\end{array}
$$

Guess: $f^{n}(x)=\frac{x}{a^{n}+b x\left(1+a+\cdots+a^{n}\right)}$
The base case is obvious. Assume it holds for $n=k$, then for $n=k+1$,
$$
f^{k \cdot H}(x)=f\left(f^{(k)}(x)\right)=\frac{\frac{x}{a^{k}+b x \frac{1-a^{k}}{1-a}}}{a+b \cdot \frac{x}{a^{k}+b x \frac{1-a^{k}}{1-a}}}=\frac{x}{a^{k+1}+b x \frac{1-a^{k+1}}{1-a}}
$$

The proposition is proved.