Example 4 Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{n}=$ $a_{n-1} b_{n}, b_{n}=\frac{b_{n-1}}{1-a_{n-1}^{2}}(n \geqslant 2), a_{1}=p, b_{1}=$ $q$, and $p, q>0, p+q=1$. Find the general terms of $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$.
Solution: When $n=1$, $a_{1}=p, b_{1}=q$.
When $n=2$,
$$
\begin{array}{l}
b_{2}=\frac{q}{1-p^{2}}=\frac{1-p}{1-p^{2}}=\frac{1}{1+p}, \\
a_{2}=a_{1} b_{2}=\frac{p}{1+p} .
\end{array}
$$

When $n=3$,
$$
\begin{array}{c}
b_{3}=\frac{\frac{q+p}{q+2 p}}{1-\frac{p^{2}}{(1+p)^{2}}}=\frac{1+p}{1+2 p}, \\
a_{3}=a_{2} b_{3}=\frac{p}{1+p} \cdot \frac{1+p}{1+2 p}=\frac{p}{1+2 p} .
\end{array}
$$

Conjecture: $a_{n}=\frac{p}{1+(n-1) p}$,
$$
b_{n}=\frac{1+(n-2) p}{1+(n-1) p} \text {. }
$$

Proof:
For $n=1,2,3$, it has been verified.
Assume for $n=k$,
$a_{k}=\frac{p}{1+(k-1) p}, b_{k}=\frac{1+(k-2) p}{1+(k-1) p}$. Then for $n=k+1$,
$b_{k+1}=\frac{b_{k}}{1-a_{k}^{2}}=\frac{\frac{1+(k-2) p}{1+(k-1) p}}{1-\left(\frac{p}{1+(k-1) p}\right)^{2}}$
$$
\begin{array}{l}
\quad=\frac{1+(k-2) p}{1+(k-1) p} \cdot \frac{[1+(k-1) p]^{2}}{(1+(k-1) p]^{2}-p^{2}} \\
\quad=\frac{1+(k-1) p}{1+k p} . \\
a_{k+1}=a_{k} b_{k+1} \\
=\frac{p}{1+(k-1) p} \cdot \frac{1+(k-1) p}{1+k p} \\
=\frac{p}{1+k p} . \\
\therefore a_{n}=\frac{p}{1+(n-1) p}, b_{n}=\frac{1+(n-2) p}{1+(n-1) p}
\end{array}
$$

For $n \in \mathbf{N}$, it holds.