$n_1,...,n_k$ pair coprime $\!\iff\! {\rm lcm}(n_1,...,n_k)=n_1...n_k$ [lcm = product for coprimes] $n_1,...,n_k$ pairwise coprime $\iff LCM(n_1,...,n_k)=n_1...n_k$
Recently, I was told this as part of a larger proof concerning direct products of groups. I am wondering why this is true. 

We notate $\,n_1,n_2,\ldots, n_k\,$ as $\,a,b,\cdots, i,j\,$ to avoid notational obfuscation.
$(\Leftarrow)\,\ \ 1< n\mid i,j\,\Rightarrow\,  a,\ldots ,i,j\, \mid\, a\cdots ij/n,\,$ so $\,{\rm lcm}(a,\ldots,i,j) \le a\cdots ij/n$
$(\Rightarrow)\,\ $ Write $\,[a,b,\ldots]:= {\rm lcm}(a,b,\ldots).\,$ We induct using  $\,\color{#c00}{\rm U} = $ LCM Universal Property
$$ a_1,\ldots,a_k\mid m\!\!\overset{\color{#c00}{\rm U\!\!}}\iff [a_1,\ldots,a_k]\mid m\qquad\qquad\qquad\qquad\qquad\qquad\  $$
Induct: $  $ base case: $\ k=2\!:\,\  (a,b)=1 \Rightarrow [a,b] = ab_{\phantom{|}}\,$ by here.
The induction step $\,k>2\,$ is as follows, using the base case and  $\color{#c00}{\rm U}$ a few times
$$\begin{align}[a,b,...,j]\mid m &\iff a,\,b,\ldots,j\ \mid m\\[.2em]
&\iff  a,[b,\ldots,j]\mid m\\[.2em]
&\iff a,\ b\ \cdots\:\  j\  \mid\, m,\ \ \ \text{by induction}\\[.2em]
&\iff a\cdot b\ \cdots\:\  j\  \mid\, m,\ \ \ \text{by base case $\,k = 2\,$ and by}
\end{align} $$
employing: $\,a\,$ coprime to $\,b,\ldots,j\iff a\,$ coprime to $\,b\cdots j,\,$ by here.
Remark $ $ Above "$\!\!\iff\!\!$" chain implies $[a,...,j]$ and $\,a\cdots j\,$ divide each other (i.e. are associate), so being $> 0$ in $\Bbb Z$ they're equal. In general domains the best we can say is that they are equal up to a unit multiple. But in many cases we can effectively unit normalize GCDs and LCMs to get equality there too (or we can ignore unit factors by working in the monoid quotient modulo the unit group).
