Example 8. In a convex $n(n \geq 4)$-sided polygon, if any three of all the diagonals do not intersect at the same point, into how many parts do they divide the $n$-sided polygon?
Method One: Draw the diagonals in sequence. When a new diagonal is drawn, if the number of intersection points with the already drawn diagonals is $m$, then this new diagonal increases the number of segments the $n$-sided polygon is divided into by $m+1$. According to this incremental pattern, the number of segments the diagonals divide the $n$-sided polygon into is: $D+P+1, D--n$-sided polygon diagonal number $\frac{n(n-3)}{2} ; P$ -- number of diagonal intersection points $C_{n}^{4}$. Therefore, $D+P+1=C_{\mathrm{n}-1}^{2}+C_{4}^{4}$.

Method Two: Suppose the diagonals divide the $n$-sided polygon into $f(n)$ parts, then there is the following recursive relationship:
$$
f(n+1)=f(n)+x+1,
$$

where $x$ is the number of segments into which the diagonals from vertex $A_{\mathrm{n}+1}$ are divided by the other diagonals. After expressing $x$ in terms of $n$, the recursive method can be applied.