Example 4 Let $a>0$. As shown in Figure 1, given the line $l: y=a x$ and the curve $C: y=x^{2}$, the x-coordinate of point $Q_{1}$ on curve $C$ is $a_{1}\left(0<a_{1}<a\right)$. From point $Q_{n}(n \geqslant 1)$ on curve $C$, draw a line parallel to the $x$-axis, intersecting line $l$ at point $P_{n+1}(n=1,2, \cdots)$. Then, from point $P_{n+1}$, draw a line parallel to the $y$-axis intersecting curve $C$ at point $Q_{n+1}$. The x-coordinates of $Q_{n}$ form the sequence $\left\{a_{n}\right\}$.

Try to find the relationship between $a_{n+1}$ and $a_{n}$, and derive the general term formula for $a_{n}$.
(2003, National College Entrance Examination (Jiangsu Province))
Notice
$$
\begin{array}{l}
Q_{n}\left(a_{n}, a_{n}^{2}\right), P_{n+1}\left(\frac{a_{n}^{2}}{a}, a_{n}^{2}\right), \\
Q_{n+1}\left(a_{n+1}, a_{n+1}^{2}\right)=\left(\frac{a_{n}^{2}}{a}, \frac{a_{n}^{4}}{a^{2}}\right) .
\end{array}
$$

Thus, $a_{n+1}=\frac{a_{n}^{2}}{a}$.
Taking the common logarithm, we get $\lg a_{n+1}=2 \lg a_{n}-\lg a$.
Therefore, $\lg a_{n+1}-\lg a=2\left(\lg a_{n}-\lg a\right)$.
Hence, $\left\{\lg a_{n}-\lg a\right\}$ is a geometric sequence with the first term $\lg a_{1}-\lg a$ and common ratio 2.
Thus, $\lg a_{n}-\lg a=\left(\lg a_{1}-\lg a\right) 2^{n-1}$, which means $a_{n}=a\left(\frac{a_{1}}{a}\right)^{2 n-1}$.