LCM is associative: $\ \text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))$ 
Definition: The least common multiple of $a.b\in\Bbb{Z}$ is the smallest  $k\in\Bbb{N}$ that is divisible by both a and b.
  

Proposition: $\text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))\tag{2}$


Proof: We want to show $\text{lcm}(\text{lcm}(a,b),c)\mid\text{lcm}(a,\text{lcm}(b,c))$ and $\text{lcm}(a,\text{lcm}(b,c))\mid \text{lcm}(\text{lcm}(a,b),c)$ such that $\text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))$.
Let $k=\text{lcm}(\text{lcm}(a,b),c)$, then $c\mid k$ and $\text{lcm}(a,b)\mid k$, but if we let $l=\text{lcm}(a,b)$, then $a\mid l\,\,\text{and}\,\, b\mid l$. Therefore $c\mid k\,\,,a\mid k\,\,\text{and}\,\, b\mid k$, but then $\text{lcm}(b,c)\mid k$. Thus $\text{lcm}(a,\text{lcm}(b,c))\mid k$ 

Question: Does this makes sense, ifso can I do the same for $\text{lcm}(\text{lcm}(a,b),c)\mid \text{lcm}(a,\text{lcm}(b,c))$?

 A: Yes, that's the idea. More succinctly, using  these universal properties of lcm and max,
$$\begin{eqnarray}\rm a,b\:\mid\, n&&\rm \iff  [\,a,b\,]\,\mid\: n\quad\ for\ \quad [\,a,b] := \ lcm(a,b)\\[.4em]
\rm a,b \le n&&\rm \iff \lceil a,b\rceil\!\le n\quad\ for\ \quad \lceil a,b\rceil := max(a,b)\end{eqnarray}$$
we obtain the following two proofs, which have precisely the same form
$$\begin{eqnarray} &&\rm[\ [\,a,b],\,c]\:\!\mid\:\! n\!\iff\! [\,a,b\,],c\,\mid\, n\!\iff\! a,b,c\,\mid \,n\!\iff\! a,[\,b,c\,]\,\mid\, n\!\iff\! [\,a,[\,b,c\,]\,]\:\!\mid\:\! n\\[.4em]
\rm  &&\rm\lceil\lceil a,b\rceil,c\rceil\! \le\! n\!\iff\! \lceil a,b\rceil,c \le\! n\!\iff\! a,b,c \le\!n\!\iff\! a,\lceil b,c\rceil\! \le n\!\iff\! \lceil a,\lceil b,c\rceil\rceil\!\le n\end{eqnarray}$$
In fact, the lcm proof transforms into the max proof if one works with exponents in prime factorizations. In the end we see that the associativity of each of these operations boils down to the associativity of "logical and", which is implicit in the notation employed in the middle terms above, i.e. $\rm\: a,b,c\mid n\:$ means $\rm\:a\mid n\:\land\: b\mid n\:\land\: c\mid n.\: $ Associating both ways yields both lcm associations.
A: Yes, you're totally right and can do the same for the other direction.
