Proof that geometric product is associative

Geometric product has nice property since it is a ring and it is associative to multiplication, which is not the case for vector cross product. But besides it is an axiom for geometric product, in the process of actually defining geometric product in a constructive way, is there a proof that it is indeed satisfy the associativity? i.e., the geometric product of a blade $A_r$ and blade $B_s$ by grade expansion of $$A_rB_s = \langle A_rB_s\rangle_{|r-s|} + ... + \langle A_rB_s\rangle_{r+s}$$ is this associative?

• Isn't it one of the three axioms for the geometric product? You can't prove an axiom! – user122283 Aug 29 '14 at 15:15
• For the main examples it is easy to see that they are associative. – Dietrich Burde Aug 29 '14 at 15:17
• The usual approach is to take the fact that the multiplication is associative as an axiom. See Chapter 1 of Hestenes and Sobczyk, Clifford Algebra to Geometric Calculus, Reidel 1984. – almagest Aug 29 '14 at 15:22
• @Sanath, you have to prove the computation rules made up for the geometric product do satisfy the axioms. Otherwise you know there is a thing called geometric product, but do not know whether it is the same thing you are calculating. – ahala Aug 29 '14 at 15:28
• @ahala You're taking as axiomatic that the geometric product is associative, so what "rules" do you think must be shown to be consistent with this associative axiom? – Muphrid Aug 29 '14 at 16:26

If you are defining it from the top down as a quotient of the tensor algebra on $V$, then it is associative because the tensor algebra is associative.