Proving that GCD of $3$ numbers is same as the GCD of GCD of $2$ numbers and the third.

Question

Here's the problem:

Prove that $(a,b,c) = ((a,b),c)$, where $(\cdot)$ denotes the GCD.

This problem looks so trivial at staring, but rigorously proving it turns out to be difficult task for me.

I know of $2$ ways of defining GCD (of, say $b$ and $c$): The least positive value of $(bx_1 + cy_1)$; or the common divisor which is divisible by every other common divisor. This definition can extend to more than $2$ numbers also.

So, I let $g$ be any common divisor of $a$, $b$ and $c$, and $G$ be the greatest of them.

Hence, matters are left to merely proving $G = ((a,b),c)$.

It is quite easily seen that $G|(a,b)$ and $G|c$. Hence, $G|((a,b),c)$. But it leaves the proof to: $((a,b),c)|G$, which I can't prove.

Perhaps it is proving: $g|((a,b),c)$, for any $g$ (which is defined above). This can lead to another proof, but I didn't find a way out!

Please help!

Hint: You can also prove that $(a,b,c)$ is the smallest positive value of $ak+bl+cm$. — WoolierThanThou, Oct 27 '20 at 13:13
Another proof just goes via an appeal to the fundamental theorem of arithmetic and comparing elementary divisors. — WoolierThanThou, Oct 27 '20 at 13:14
@WoolierThanThou But how can I be sure that I proved the original problem? It states equality. — Book Of Flames, Oct 27 '20 at 13:15
@Spectre I am newbie to number theory, so I know only formal basic theorems, not any strategy. — Book Of Flames, Oct 27 '20 at 13:18
@BookOfFlames, I am nowhere in the sea of number theory, but I am hoping o get through it :) — Spectre, Oct 27 '20 at 13:18
@BookOfFlames Can you rephrase your question? I'm saying the statement follows quite automatically once you've proven my first hint (or used the second one if you prefer). Where are you worried that there might be a problem of equality? — WoolierThanThou, Oct 27 '20 at 13:32
@WoolierThanThou I didn't quite get that. You can add an answer to add more details. — Book Of Flames, Oct 27 '20 at 13:35

Proving that GCD of $3$ numbers is same as the GCD of GCD of $2$ numbers and the third.

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