Example 9 In a convex $n$-sided polygon, drawing $(n-3)$ non-intersecting diagonals, how many ways are there to do this?
Solution: It is easy to know that a convex $n$-sided polygon has at most $n-3$ non-intersecting diagonals. Let the number of different connection methods be $a_{n}$. Obviously, $a_{4}=2$, and it is agreed that $a_{3}=a_{2}=1$. For the convex $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, suppose the diagonal $A_{1} A_{k}$ passes through $A_{1}$, and there are no $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{k-1}$. Then, the $(k-1)$-sided polygon $A_{2} A_{3} \cdots A_{k}$ can have $k-4$ non-intersecting diagonals in $a_{k-1}$ ways, and the $(n-k+2)$-sided polygon $A_{b} A_{k \cdot 1} \cdots A_{n} A_{1}$ can have $n-k-1$ non-intersecting diagonals in $a_{n-k+2}$ ways, totaling $a_{k-1} a_{n-k+1}$ ways. Each of these (adding $A_{2} A_{k}$ and $A_{1} A_{k}$) becomes one of $a_{n}$. By setting $k=3,4, \cdots, n$, we get the recurrence relation
$$
\begin{aligned}
a_{n}= & a_{2} a_{n-1}+a_{3} a_{n-2}+a_{4} a_{n-3}+\cdots+ \\
& a_{n-1} a_{2} .
\end{aligned}
$$
(The last term corresponds to $k=n$, indicating no diagonal passing through $A_{1}$)
Establish the generating function
$$
\begin{aligned}
F(x)= & a_{2}+a_{3} x+a_{4} x^{2}+\cdots+a_{n} x^{n-2} \\
& +\cdots .
\end{aligned}
$$

Then $F^{2}(x)$
$$
\begin{aligned}
= & \left(a_{2}+a_{3} x+\cdots\right)\left(a_{2}+a_{3} x+\cdots\right) \\
= & a_{2}^{2}+\left(a_{2} a_{3}+a_{3} a_{2}\right) x+\cdots+\left(a_{2} a_{n-1}\right. \\
& \left.+a_{3} a_{n-2}+\cdots+a_{n-1} a_{2}\right) x^{n-3}+\cdots \\
= & a_{3}+a_{4} x+a_{5} x^{2}+\cdots \\
= & \frac{F(x)-1}{x} .
\end{aligned}
$$

Solving, we get $F(x)=\frac{1 \pm \sqrt{1-4 x}}{2 x}$.
By the binomial theorem, we have
$$
\begin{aligned}
& (1-4 x)^{\frac{1}{2}} \\
= & 1+\sum_{k=1}^{n} \frac{\frac{1}{2}\left(\frac{1}{2}-1\right) \cdots\left(\frac{1}{2}-k+1\right)}{k!} . \\
& (-4)^{k} \\
= & 1-\sum_{k=1}^{\infty} \frac{2^{k} \cdot 1 \cdot 3 \cdot 5 \cdots(2 k-3)}{k!} x^{k} .
\end{aligned}
$$

Since $F(0)=a_{2}=1$, we take the negative sign in equation (1) to get
$$
\begin{array}{l}
F(x) \\
=\sum_{k=1}^{\infty} \frac{2^{k-1} \cdot 1 \cdot 3 \cdots(2 k-3)}{k!} x^{k-1} .
\end{array}
$$

Thus,
$$
\begin{array}{l}
a_{n}=\frac{2^{n-2} \cdot 1 \cdot 3 \cdots(2 n-5)}{(n-1)!} \\
=\frac{(2 n-4)!}{(n-1)!(n-2)!} .
\end{array}
$$