Proof

Let $a,b,c\in\mathbb Z$ be integers.
The commutativity of $\gcd$ follows immediatly from the definition of the greatest common divisor because of the associativity of conjunction; i.e. since $d\mid a\wedge d\mid b$ if and only if $d\mid b\wedge d\mid a,$ the sets $D_{a,b}$ and $D_{b,a}$ are equal and have the same maximum.
The associativity of $\gcd$ can be understood as follows:
- If $t$ is a divisor of both, $t\mid \gcd(a,b)$ and $t\mid c,$ then $t\mid \gcd(\gcd(a,b),c)).$ Moreover, $t\mid a$ and $t\mid b.$
- Therefore, $t$ is also divisor of both, $t\mid a$ and $t\mid \gcd(b,c).$ Thus $t$ divides $t\mid \gcd(a,\gcd(b,c))$
- Since this reasoning holds for any common divisor of $t\mid a,$ $t\mid b,$ and $t\mid c,$ it follows $\gcd(\gcd(a,b),c))=\gcd(a,\gcd(b,c)).$
Since $\gcd$ is both, commutative and associative, the notation $\gcd(a,b,c)$ makes sense.
By induction over $k,$ the formula $\gcd(n_1,\ldots,n_k)=\gcd(\gcd(n_1,\ldots,n_{k-1}),n_k)$ follows.
∎

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Scheid Harald: "Zahlentheorie", Spektrum Akademischer Verlag, 2003, 3rd Edition
Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927