Proof of the Associative property of the Least common multiple How to prove associative property of the Least common multiple (lcm). i.e,

$$
\mathcal{lcm}(a,b,c)=\mathcal{lcm}(a,\mathcal{lcm}(b,c))=\mathcal{lcm}(\mathcal{lcm}(a,b),c)
$$

My Approach:
Thanx to @Mastrem for the hint.
Using the fundamental theorem of Arithmetic,
$$
a=2^{a_{2}}.3^{a_{3}}.5^{a_{5}}......=\prod_{p} p^{a_p}\quad b=2^{b_{2}}.3^{b_{3}}.5^{b_{5}}......=\prod_{p} p^{b_p}\\
lcm(a,b)=\prod_p p^{\max(a_p,b_p)}
$$
Using that,
$$
lcm(a,b,c)=lcm(a,lcm(b,c))=lcm(a,\prod_p p^{\max(b_p,c_p)})=\prod_p p^{\max(a_p,\prod_p p^{\max(b_p,c_p)})}
$$
Similarly,
$$
lcm(a,b,c)=lcm(lcm(a,b),c)=lcm(\prod_p p^{\max(a_p,b_p)},c)=\prod_p p^{\max(\prod_p p^{\max(a_p,b_p)},c_p)}
$$
How do I proceed further and equate the RHS of both the equations and complete the proof ?
 A: Thanx to @Guy Fsone, @Mastrem for their hints.
Using the fundamental theorem of arithmetic
$$
a=p_1^{a_1}.p_2^{a_2}.p_3^{a_3}.....p_k^{a_k}=\prod_ip_i^{a_i}\\b=p_1^{b_1}.p_2^{b_2}.p_3^{b_3}.....p_k^{b_k}=\prod_ip_i^{b_i}\\c=p_1^{c_1}.p_2^{c_2}.p_3^{c_3}.....p_k^{c_k}=\prod_ip_i^{c_i}
$$
where $p_{i}$ are prime numbers, i.e, $p_1=2, p_2=3, p_3=5, p_4=7, ....$so on and 
$$
lcm(b,c)=p_1^{max(b_1,c_1)}.p_2^{max(b_2,c_2)}.p_3^{max(b_3,c_3)}.....p_k^{max(b_k,c_k)}=\prod_ip_i^{max(b_i,c_i)}\\
lcm(a,b)=p_1^{max(a_1,b_1)}.p_2^{max(a_2,b_2)}.p_3^{max(a_3,b_3)}.....p_k^{max(a_k,b_k)}=\prod_ip_i^{max(a_i,b_i)}
$$
and using the associative property of max.
$$
max(a,b,c)=max(a,max(b,c))=max(max(a,b),c)
$$
Now,
$$
lcm(a,b,c)=lcm(a,lcm(b,c))=lcm\Big(a,p_1^{max(b_1,c_1)}.p_2^{max(b_2,c_2)}.p_3^{max(b_3,c_3)}.....p_k^{max(b_k,c_k)}\Big)=lcm\Big(p_1^{a_1}.p_2^{a_2}.p_3^{a_3}.....p_k^{a_k},p_1^{max(b_1,c_1)}.p_2^{max(b_2,c_2)}.p_3^{max(b_3,c_3)}.....p_k^{max(b_k,c_k)}\Big)=p_1^{max\big(a_1,max(b_1,c_1)\big)}.p_2^{max\big(a_2,max(b_2,c_2)\big)}.p_3^{max\big(a_3,max(b_3,c_3)\big)}.....p_k^{max\big(a_k,max(b_k,c_k)\big)}\\=p_1^{max(a_1,b_1,c_1)}.p_2^{max(a_2,b_2,c_2)}.p_3^{max(a_3,b_3,c_3)}.....p_k^{max(a_k,b_k,c_k)}=\prod_{i}p_{i}^{max(a_i,b_i,c_i)}
$$
similarly,
$$
lcm(lcm(a,b),c)=lcm\Big(p_1^{max(a_1,b_1)}.p_2^{max(a_2,b_2)}.p_3^{max(a_3,b_3)}.....p_k^{max(a_k,b_k)},c\Big)=lcm\Big(p_1^{max(a_1,b_1)}.p_2^{max(a_2,b_2)}.p_3^{max(a_3,b_3)}.....p_k^{max(a_k,b_k)},p_1^{c_1}.p_2^{c_2}.p_3^{c_3}.....p_k^{c_k},\Big)=p_1^{max\big(max(a_1,b_1),c_1\big)}.p_2^{max\big(max(a_2,b_2),c_2\big)}.p_3^{max\big(max(a_3,b_3).c_3\big)}.....p_k^{max\big(max(a_k,b_k),c_k\big)}\\=p_1^{max(a_1,b_1,c_1)}.p_2^{max(a_2,b_2,c_2)}.p_3^{max(a_3,b_3,c_3)}.....p_k^{max(a_k,b_k,c_k)}=\prod_{i}p_{i}^{max(a_i,b_i,c_i)}
$$
Since,
$lcm(a,lcm(b,c))=\prod_{i}p_{i}^{max(a_i,b_i,c_i)}=lcm(lcm(a,b),c)$, the associative property of the lcm is thus proved.
A: Hint: For every prime $p$ and natural number $n$, let $e_p(n)$ be the exponent of the highest power of $p$ dividing $n$, so 
$$n=\prod_{p\text{ prime}}p^{e_p(n)}$$
and use the fact that:
$$\operatorname{lcm}(a,b)=\prod_{p\text{ prime}}p^{\max(e_p(a),e_p(b))}$$
A: If you view numbers as a sequence of prime factor exponents (which by the fundamental theorem of arithmetic is unique), then $\text{lcm}$ on two numbers is simply taking the maximum of each prime factor exponent.
Since the $\max$ function is associative, we automatically prove that $\text{lcm}$ is too. And the $\max(a, b, c)$ of three numbers is also equal to $\max(a, \max(b, c))$ by transitivity.
