By definition, $\operatorname{LCM}\left\{b_{1}, b_{2}, \cdots b_{m}\right\}$ is the smallest positive integer that can be divided by $b_{1}, \cdots, b_{m}$. If we set their factorization as
$$\left\{\begin{array}{l} b_{1}=p_{1}^{a_{1}^{(1)}} \cdots \cdots p_{n}^{a_{n}^{(1)}}, a_{1}^{(1)} \geqslant 0, \cdots, a_{n}^{(1)} \geqslant 0, \\ \cdots \cdots \cdots \cdots \\ b_{m}=p_{1}^{a_{1}^{(m)}} \cdots \cdots p_{n}^{a_{n}^{(m)}}, a_{1}^{(m)} \geqslant 0, \cdots, a_{n}^{(m)} \geqslant 0, \end{array}\right.$$

then the following formula holds:
$$\operatorname{LCM}\left\{b_{1}, \cdots, b_{m}\right\}=p_{1}^{a_{1}} \cdots p_{n}^{a_{n}}$$

where
$$a_{j}=\max \left\{a_{j}^{(1)}, \cdots, a_{j}^{(m)}\right\}, j=1, \cdots, n$$

By the result of Example 2, we have
$$\begin{aligned} a_{j}= & \sum_{i} a_{j}^{(i)}-\sum_{i_{1} i_{2}} \min \left\{a_{j}^{\left(i_{1}\right)}, a_{j}^{(i 2)}\right\}+-\cdots \\ & +(-1)^{m+1} \min \left\{a_{j}^{(1)}, \cdots, a_{j}^{(m)}\right\}, j=1, \cdots, n \end{aligned}$$

Also, note that
$$\left(b_{i_{1}}, \cdots, b_{i_{s}}\right)=p_{1}^{c_{1}} \cdots p_{n}^{c_{n}}$$

where
$$c_{j}=\min \left\{a_{j}^{\left(i_{j}\right)}, \cdots, a_{j}^{\left(i_{s}\right)}\right\}, j=1, \cdots, n$$

We can easily see that

This is exactly what we need to prove.
$$\begin{array}{l} p_{1}^{a_{1}} \cdots p_{n}^{a_{n}}=p_{1}^{\sum_{i} a_{1}^{(i)}} \cdots p_{n}^{\sum_{i} a_{n}^{(i)}} \cdot p_{1}^{-\sum \min \left\{a_{1}^{\left(i_{1}\right)} \cdot a_{1}^{\left(i_{2}\right)}\right\}} \cdots p_{n}^{-\sum \min \left\{a_{n}^{\left(i_{1}\right)} \cdot a_{n}^{\left(i_{2}\right)}\right\}} \end{array}$$
$$\begin{array}{l} =\left(b_{1} \cdots b_{m}\right) \cdot\left(b_{1}, b_{2}\right)^{-1} \cdots\left(b_{m-1}, b_{m}\right)^{-1} \\ \cdots\left(b_{1}, b_{2}, \cdots, b_{m}\right)^{(-1)^{m+1}}, \end{array}$$