TITLE: Concise proof that every common divisor divides GCD without Bezout's identity?
QUESTION [20 upvotes]: In the integers, it follows almost immediately from the division theorem and the fact that $a | x,y \implies a | ux + vy$ for any $u, v \in \mathbb{Z}$ that the least common multiple of $a$ and $b$ divides any other common multiple.
In contrast, proving $e|a,b \implies e|gcd(a,b)$ seems to be more difficult. In Elementary Number Theory by Jones & Jones, they do not try to prove this fact until establishing Bezout's identity. This Wikipedia page has a proof without Bezout's identity, but it is convoluted to my eyes.
I tried my hand at it, and what I got seems no cleaner:
Proposition: If $e | a,b$, then $e | gcd(a,b)$.
Proof: Let $d = gcd(a,b)$. Then if $e \nmid d$, by the division theorem there's some $q$ and $c$ such that $d = qe + c$ with $0 < c < r$.
We have $a = k_1 d$ and $b = k_2 d$, so by substituting we obtain $a = k_1 (qe + c)$ and $b = k_2 (qe + c)$. Since $e$ divides both $a$ and $b$, it must divide both $k_1 c$ and $k_2 c$ as well. This implies that both $k_1 c$ and $k_2 c$ are common multiples of $c$ and $r$.
Now let $l = lcm(r, c)$. $l$ divides both $k_1 c$ and $k_2 c$. Since $l = \phi c$ for some $\phi$, we have $\phi | k_1, k_2$, so $d \phi | a, b$.
But we must have $\phi > 1$ otherwise $l = c$, implying $r | c$, which could not be the case since $c < r$. So $d \phi$ is a common divisor greater than $d$, which is a contradiction. $\Box$
Question: Is there a cleaner proof I'm missing, or is this seemingly elementary proposition just not very easy to prove without using Bezout's identity?

REPLY [2 votes]: You might also use the fact that each natural number can be uniquely represented
as a product of prime powers to prove this fact without Bezout's Theorem.
This is called Fundamental Theorem of Arithmetic
For example  d=p1^d1*..*pn^dn where p1 to pn are primes.
The numbers p1 to pn are called then prime factors of d
Idea of the proof argument:
Since gcd(a,b) divides a and b each prime factor of gcd(a,b) has to be
a common prime factor of a and b.
For the same reason each prime factor of d has to be a common prime factor of 
a and b.
And since gcd(a,b) is the greatest common divisor of a,b(by defnition) ,each prime factor of d has to
be then a prime factor of gcd(a,b), for if there would be at least one prime factor p of 
d which isn't a prime factor of gcd(a,b) ,then the product gcd(a,b)*p would be again
a common divisor of both a and b ,greater than gcd(a,b) ,which would be a contradiction.