Example 11 In a convex $n$-sided polygon, by non-intersecting diagonals inside the $n$-sided polygon, divide the $n$-sided polygon into non-overlapping triangles. Please express the number of different methods using a recursive formula.
For a polygon $A_{1} A_{2} \cdots A_{n-1} A_{n}$, let the number of valid constructions be $a_{n}$.
Now consider the recursive relationship based on the line segment $A_{1} A_{n}$.
Construct $\triangle A_{1} A_{k} A_{n}(1<k<n)$, then the original convex $n$-gon
is divided into three parts, namely
$\triangle A_{1} A_{k} A_{n}$, the convex polygon
$A_{1} A_{2} \cdots A_{k}$, and the convex polygon
$A_{k} A_{k+1} \cdots A_{n}$ (as shown in Figure 4). By
the multiplication principle, for $A_{k}$, there are $a_{k} a_{n-k+1}$ ways to construct. Let $k$
range from $2,3, \cdots, n-1$, we get
$$
a_{n}=\sum_{k=2}^{n-1} a_{k} a_{n-k+1} .
$$

Note, the initial value should be $a_{3}=1$.