# How does multiplicative associativity and commutativity justify $\prod {a_i}^{b_i}$?

This is trivial, but what would be a simple sentence describing this property?

For example, let $R$ be an algebraic structure in which multiplicative commutativity and associativity hold. And let $\{a_1,...,a_n\}$ be a finite subset of $R$ and $\{b_1,...,b_n\}$ be a finite subset of $\omega$.

Then, no matter which order of multiplication taken, the product should be equal, so that we can write this prouct as $\prod {a_i}^{b_i}$ and this can only be defined if order of multiplication taken doesn't matter.

What is a sentence or theorem states that "Order of mulplication doesn't matter"?

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You yourself mentioned multiplicative commutativity. Isn't that the name that you want? – EuYu Feb 26 '13 at 5:33
The definition for multipliative commutativity is [$\forall a,b\in R, ab=ba$]. There should be some steps to prove whether ((ab)(bc))(ca)=((a(bb))(cc))a from the definition. – Jj- Feb 26 '13 at 5:42
If the index set is ordered then one doesn't need commutativity for the notation to be well-defined. Then it is simply generalized (n-ary) associativity. – Math Gems Feb 26 '13 at 5:46
@MathGems Thank you and I like the name of it; generalized associativity. – Jj- Feb 26 '13 at 6:16