Example 2 Given the quadratic equation in $x$
$x^{2}+n^{2} x+n-1=0(n$ is a natural number, $n \geqslant 2$ ). When $n=2$, the two roots of this equation are denoted as $\alpha_{2}$ and $\beta_{2}$; when $n=3$, the two roots are denoted as $\alpha_{3}$ and $\beta_{3}$; $\cdots \cdots$ when $n=2006$, the two roots are denoted as $\alpha_{2006}$ and $\beta_{2006}$. Find
$$
\begin{array}{l}
\frac{1}{\left(\alpha_{2}-1\right)\left(\beta_{2}-1\right)}+\frac{1}{\left(\alpha_{3}-1\right)\left(\beta_{3}-1\right)}+\cdots+ \\
\frac{1}{\left(\alpha_{2006}-1\right)\left(\beta_{2006}-1\right)}
\end{array}
$$

the value of the above expression.
Explanation: From the relationship between roots and coefficients, we have
$$
\begin{array}{l}
\alpha_{n}+\beta_{n}=-n^{2}, \alpha_{n} \beta_{n}=n-1 . \\
\text { Then }\left(\alpha_{n}-1\right)\left(\beta_{n}-1\right) \\
=\alpha_{n} \beta_{n}-\left(\alpha_{n}+\beta_{n}\right)+1 \\
=n-1+n^{2}+1 \\
=n(n+1) .
\end{array}
$$

When $n=2,3, \cdots, 2006$,
$$
\begin{array}{l}
\text { Original expression }=\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{2006 \times 2007} \\
=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots+\left(\frac{1}{2006}-\frac{1}{2007}\right) \\
=\frac{1}{2}-\frac{1}{2007}=\frac{2005}{4014} .
\end{array}
$$