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This question is inspired by the other question which asked for a proof that $i^i$ is a real number.

Many calculators when asked for $0^0$ return 1. I asked a mathematician how to prove that but he replied that it is impossible and that one only can postulate this by a convention.

So I wonder whether he is right? Is $0^0=1$ axiom really independent of all other axioms defining standard complex numbers and exponentiation?

UPDATE

It seems that no response so far tried to provide an answer to my clear question, that is whether $0^0=1$ is an independent axiom or not. Most answers are trying to defend particular values for $0^0$ which the authors prefer with some indirect or abstract argumentation, mostly involving taking limits or citing practical purpose.

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    Is there a standard axiomatization of "standard complex numbers and exponentiation"?2012-10-27
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    Many calculators return 1? Really? I have a TI-60X which I have had for around 15+ years and it returns an error. And, my TI-84 which is several years newer still returns error. So, I am guessing TI calculators probably all return errors. Can you give an example of one that returns 1?2012-10-27
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    @Graphth Windows calculator, for example.2012-10-27
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    @Anixx Any other examples? I would imagine any HP or Casio or TI calculator would give error. Can any one confirm?2012-10-27
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    This is, at most, a definition.2012-10-27
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    @Graphth Testing on all my calculators: [results](http://pastebin.com/f1F8hn3b).2012-10-28
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    @Graphth Gnome calculator on Linux. Interestingly, the KDE calculater for a while returned 0 while Gnome calculator returns 1. The KDE calculator has been fixed since following a bugreport.2012-10-28
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    Read Arturo's answer to the this question: http://math.stackexchange.com/questions/11150/zero-to-zero-power2012-11-04
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    @JavaMan thanks, the author claims that this leads to certain exceptions in limits laws so I asked for some examples in the comment.2012-11-05
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    @anixx: Arturo has not been an active member for a while (see his user page), so leaving him a comment will not yield immediate results.2012-11-05
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    @JavaMan in that case his answer seems incomplete.2012-11-05

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