Prove Proposition (2.23).
Proposition (2.23) is
Let $a_1, a_2, a_3, \ldots , a_n$ be non-zero integers, then $[a_1, a_2, a_3, \ldots , a_{n−1}, a_n] = [[a_1, a_2, a_3, \ldots , a_{n−1}] , a_n]$.
We'll use abbreviations $A$ and $B$
$$\begin{align} A & = a_1, a_2, a_3, \ldots , a_{n−1}, a_n \\ \\ B & = a_1, a_2, a_3, \ldots , a_{n−1}\end{align} $$
We need to show
$$ [A] = [[B], a_n] $$
Let the prime decomposition of $a_1$ be
$$ p_1^{\alpha(a_1)_1} \times p_2^{\alpha(a_1)_2} \times \ldots $$
Let the prime decomposition of $a_2$ be
$$ p_1^{\alpha(a_2)_1} \times p_2^{\alpha(a_2)_2} \times \ldots $$
And the prime decomposition of $a_n$ be
$$ p_1^{\alpha(a_n)_1} \times p_2^{\alpha(a_n)_2} \times \ldots $$
The LCM $[B]$ is therefore
$$ p_1^{\max(\alpha(a_1)_1, \alpha(a_2)_1, \alpha(a_3)_1, \ldots , \alpha(a_{n-1})_1} $$
$$\times $$
$$ p_2^{\max(\alpha(a_1)_2, \alpha(a_2)_2, \alpha(a_3)_2, \ldots , \alpha(a_{n-1})_2}$$
$$ \times $$
$$ \ldots $$
The LCM $[[B], a_n]$ is
$$ p_1^{\max(\alpha(a_1)_1, \alpha(a_2)_1, \alpha(a_3)_1, \ldots , \alpha(a_{n-1})_1, \textcolor{red}{\alpha(a_n)_1}} $$
$$\times $$
$$ p_2^{\max(\alpha(a_1)_2, \alpha(a_2)_2, \alpha(a_3)_2, \ldots , \alpha(a_{n-1})_2, \textcolor{red}{\alpha(a_n)_1}}$$
$$ \times $$
$$ \ldots $$
This is the LCM $[A]$.
Thus we have shown $ [A] = [[B], a_n] $ by showing the prime decompositions are the same.