If $a_1, a_2, ..., a_m$ each divide $n$, prove that $lcm(a_1,a_2, ..., a_m) | n$. If $a_1, a_2, ..., a_m$ each divide $n$, prove that $lcm(a_1,a_2, ..., a_m) | n$.  
I see that essentially says show that the lcm of factors of $n$ divide $n$ and this makes sense intuitively, but I can't seem to show it.  I can see that if $\gcd(a_i,a_j)=1$ then $a_ia_j = lcm(a_i,a_j) | a$, but when $\gcd(a_i,a_j) \neq 1$ I get stuck.
 A: Let the prime factorization of $n$ be
$$n = \prod_{j=1}^q p_j^{v_j}.$$
Now since $a_k|n$ we have
$$a_k = \prod_{j=1}^q p_j^{w_{j,k}}$$
where $w_{j,k}\le v_j.$ We also have that
$$L = \mathrm{lcm}(a_1, a_2,\ldots, a_m)
= \prod_{j=1}^q p_j^{\max_k w_{j,k}}.$$
Note however that $\max_k w_{j,k} \le v_j$ and hence $L|n$ and we have
the claim.
A: That $\ a_1,\ldots,a_k\mid m\,\Rightarrow\,{\rm lcm}(a_1,\ldots,a_k)\mid m\ $ is a prototypical  Euclidean descent.
The set $M$ of all positive common multiples of all $\,a_i$ is closed under positive subtraction, i.e. $\,m> n\in M$ $\Rightarrow$ $\,a_i\mid m,n\,\Rightarrow\, a_i\mid m\!-\!n\,\Rightarrow\,m\!-\!n\in M.\,$ Thus, by induction, we deduce that $\,M\,$ is closed under mod, i.e. remainder, since it arises by repeated subtraction, i.e. $\ m\ {\rm mod}\ n\, =\, m-qn = ((m-n)-n)-\cdots-n.\,$ Thus the least $\,\ell\in M\,$ divides every  $\,m\in M,\,$ else $\ 0\ne m\ {\rm mod}\ \ell\ $ is in $\,M\,$ and smaller than $\,\ell,\,$ contra minimality of $\,\ell.$
