Example 7 Given an integer $n \geqslant 2$, for any coprime positive integers $a_{1}, a_{2}, \cdots, a_{n}$, let
$$
A=a_{1}+a_{2}+\cdots+a_{n} .
$$

Let the greatest common divisor of $A$ and $a_{i}(i=1,2, \cdots, n)$ be $d_{i}$; the greatest common divisor of the remaining $n-1$ numbers after removing $a_{i}$ from $a_{1}, a_{2}, \cdots, a_{n}$ be $D_{i}$. Find the minimum value of $\prod_{i=1}^{n} \frac{A-a_{i}}{d_{i} D_{i}}$.
[5]
Consider $D_{1}=\left(a_{2}, a_{3}, \cdots, a_{n}\right)$ and $d_{2}=\left(a_{2}, A\right)=\left(a_{2}, a_{1}+a_{2}+\cdots+a_{n}\right)$. Let $\left(D_{1}, d_{2}\right)=d$. Then $d\left|a_{2}, d\right| a_{3}, \cdots, d\left|a_{n}, d\right| A$. Hence $d \mid a_{1}$. Therefore, $d \mid\left(a_{1}, a_{2}, \cdots, a_{n}\right)$.
When $a_{1}, a_{2}, \cdots, a_{n}$ are pairwise coprime, we have $d=1$.
Notice that, $D_{1}\left|a_{2}, d_{2}\right| a_{2}$, and $\left(D_{1}, d_{2}\right)=1$.
Thus, $D_{1} d_{2} \mid a_{2} \Rightarrow D_{1} d_{2} \leqslant a_{2}$.
Similarly, $D_{2} d_{3} \leqslant a_{3}, \cdots, D_{n} d_{1} \leqslant a_{1}$.
Hence $\prod_{i=1}^{n} d_{i} D_{i}=\left(D_{1} d_{2}\right)\left(D_{2} d_{3}\right) \cdots\left(D_{n} d_{1}\right)$
$\leqslant a_{2} a_{3} \cdots a_{n} a_{1}=\prod_{i=1}^{n} a_{i}$.
Considering,
$$
\begin{array}{l}
\prod_{i=1}^{n}\left(A-a_{i}\right)=\prod_{i=1}^{n}\left(\sum_{j \neq i} a_{j}\right) \\
\geqslant \prod_{i=1}^{n}\left[(n-1)\left(\prod_{j \neq i} a_{j}\right)^{\frac{1}{n-1}}\right] \\
=(n-1)^{n} \prod_{i=1}^{n} a_{i} .
\end{array}
$$

From equations (1) and (2), it is easy to see that $\prod_{i=1}^{n} \frac{A-a_{i}}{d_{i} D_{i}} \geqslant(n-1)^{n}$.
On the other hand, when $a_{1}=a_{2}=\cdots=a_{n}=1$,
$$
\prod_{i=1}^{n} \frac{A-a_{i}}{d_{i} D_{i}}=(n-1)^{n} \text {. }
$$

In summary, the minimum value of $\prod_{i=1}^{n} \frac{A-a_{i}}{d_{i} D_{i}}$ is $(n-1)^{n}$.