Prove $(a,b,c)=((a,b),(a,c))$ The notation is for the greatest common divisor.  I know that 
$$(a,b,c)=((a,b),c)=((a,c),b)=(a,(b,c))$$
Suppose $g=(a,b,c)$.  Then $g\mid a,b,c$.  Also, $g\mid(a,b),c$ and $g\mid(a,c),b$.  Thus there exist integers $k,m$ such that
$$(a,b)=gk, (a,c)=gm$$
Then
$$((a,b),(a,c))=(gk,gm)=g(k,m)$$
Therefore, since $(k,m)=1$ (otherwise $g$ would not be the greatest common divisor of $(a,b)$ and $(a,c)$), $g=((a,b),(a,c))$.
Is this proof okay?

You can use just the one identity you know along with symmetry, and  nothing else, to simplify
$$ ((a,b),(a,c)) = ((a,b),a,c) = (((a,b),a), c) = ((a,a,b),c) = (((a,a),b),c) = ((a,a),b,c)$$
In fact, the associative identity for a binary operator $((a,b),c) = (a,(b,c))$ -- or in its more common expression for operators with infix notation $(a \cdot b) \cdot c = a \cdot(b\cdot c)$ -- is enough to prove that if you operate on an arbitrary number of things, it doesn't matter how you group them: e.g. if you know that binary gcds are associative, then it automatically follows, e.g., that
$$ ((a,(b,c)),d) = ((a,b),(c,d)) $$
without any fuss. And it means that extending the operator to any positive, finite number of terms is unambiguous: if I write
$$ (a,b,c,d) $$
then it doesn't matter how I group the terms into pairwise gcds, I have to get the same value.
If the operator is also symmetric (more commonly called "commutative" when talking about operations, but "symmetric" -- that is, $(a,b) = (b,a)$ or for infix operations $a\cdot b = b \cdot a$ -- that means you can rearrange the terms arbitrarily.
And thus, I immediately know that it makes sense to just say
$$ ((a,b),(a,c)) = (a,a,b,c) $$
and furthermore, if I instead write it as
$$ ((a,b),(a,c)) = ((a,a),(b,c))$$
this could be proven without ever resorting to ternary or quaternary gcds, just by using the associative and commutative laws.
