On Friday, October 4, 2013 7:45:20 PM UTC-4, fom wrote:
> > Consider the phrase: > > > > "Notwithstanding combinatorial analogies and > > arguments of convenience..." > > > > You have not *proven* that your criticism is > > not an "argument of convenience". > > > > The burden of proof has been "conveniently" > > placed upon all contrary opinions.
The fact is, like 0/0, any value with work for 0^0. If you also want to assume that x^(y+z) = x^y * x^z for all x,y,z in N, then you could narrow it down to either 0 or 1. As far as I know, there is no compelling logical reason to choose any one particular value. IMHO, none have been presented here.