Example 8 Simplify $\tan \alpha \tan 2 \alpha+\tan 2 \alpha \tan 3 \alpha+\cdots+\tan (n-1) \alpha \tan n \alpha$.
$$
\begin{array}{l}
\text { Solution: From } \tan \alpha=\tan [n \alpha-(n-1) \alpha]=\frac{\tan n \alpha-\tan (n-1) \alpha}{1+\tan n \alpha \tan (n-1) \alpha} \text {, we transform to get } \\
\begin{aligned}
\tan n \alpha \tan (n-1) \alpha=\frac{\tan n \alpha-\tan (n-1) \alpha}{\tan \alpha}-1, \text { where } n=2,3,4, \cdots \\
\text { Therefore, the original expression }=\left(\frac{\tan 2 \alpha-\tan \alpha}{\tan \alpha}-1\right)+\left(\frac{\tan 3 \alpha-\tan 2 \alpha}{\tan \alpha}-1\right)+\cdots \\
\quad+\left(\frac{\tan n \alpha-\tan (n-1) \alpha}{\tan \alpha}-1\right)=\frac{\tan n \alpha}{\tan \alpha}-n .
\end{aligned}
\end{array}
$$