[g_]

What do you mean by 'is defined by group operation'? Reals/integers with multiplication don't form a group.

'Multiplying 0 by itself 0 times' is what Wikipedia calls 'empty product'. For me is is good intuition, but that's a matter of taste.

AFAIK the usual convention is:

* in contexts where the exponent is varying continuously and a is a real number, a^b is undefined for a=b=0

* in contexts where the exponent is varying discretely (as an integer/natural number), a^b = 1 for b=0 and any a

Haskell quite nicely distinguishes between these two:

(^) :: (Num a, Integral b) => a -> b -> a

(* * ) :: (Floating a) => a -> a -> a

but it still gives 0 * * 0 and 0^0 as 1. (There shouldn't be spaces between asterisks, HN treats them as italics)

[profquail]

Reals form a group under multiplication. Integers don't, but they do form a commutative ring (addition, negation, multiplication).

[jibiki]

> Reals form a group under multiplication.

I think his point was that 0 has no inverse. So in the literal sense, the reals are not a group under multiplication (although obviously, when somebody says "the reals are a group under multiplication", one usually interprets it as "R\{0} is a group under multiplication".)

This is vaguely relevant to the issue at hand because we are doing 0 to the 0.