Example 5 A convex $n$-sided polygon $(n \geqslant 4)$, none of its three diagonals intersect at a single point, find the total number of all diagonal intersection points.
Solution: Let $a_{n}$ be the total number of diagonal intersection points in a convex $n$-gon. First, determine the initial value; it is easy to know that $a_{4}=1$.

When $n \geqslant 5$, the intersection points of all diagonals in the convex $n$-gon $A_{1} A_{2} \cdots A_{n-1} A_{n}$ can be divided into two parts: one part is the intersection points of all diagonals in the convex $(n-1)$-gon $A_{1} A_{2} \cdots A_{n-1}$, which is $a_{n-1}$; the other part is the intersection points of the diagonals drawn from $A_{n}$ with other diagonals. This can be illustrated by drawing the diagonal $A_{n} A_{k} (k=2,3, \cdots, n-2)$. On one side of $A_{n} A_{k}$, there are vertices $A_{1}, A_{2}, \cdots, A_{k-1}$, and on the other side, there are $n-k-1$ vertices $A_{k+1}, A_{k+2}, \cdots, A_{n-1}$. By selecting one point from each side and connecting them, a diagonal is formed that intersects $A_{n} A_{k}$ at one point. Therefore, the number of intersection points of $A_{n} A_{k}$ with other diagonals is $(k-1)(n-k-1)$. Taking $k=2,3, \cdots, n-2$, the total number of intersection points in the second part is
$$
\begin{array}{l}
1 \cdot(n-3)+2 \cdot(n-4)+\cdots+(n-4) \cdot 2+(n-3) \cdot 1 \\
=\frac{(n-1)(n-2)(n-3)}{6}=C_{n-1}^{3},
\end{array}
$$

Thus, we obtain the recurrence relation $a_{n}=a_{n-1}+C_{n-1}^{3} \quad(n=5,6, \cdots)$,
So $a_{n}=a_{4}+\sum_{k=5}^{n} C_{k-1}^{3}=1+C_{4}^{3}+C_{5}^{3}+\cdots+C_{n-1}^{3}=C_{n}^{4}$.
Therefore, the total number of diagonal intersection points is $C_{n}^{4}$.