I know $1^\omega$ is $1$ because for each $x < \omega$: $1^x = 1$.
But what about $0^\omega$? To be able to answer this, first, I need to know what $0^0$ is.
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1See this post – May 25 '18 at 16:01
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@LearningMathematics: Pretty much all of the suggestions in that thread are irrelevant to this question. – Asaf Karagila May 25 '18 at 16:20
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Well, $0^0$ is $1$. This follows from any number of definitions, at least if you want them to be "reasonable" and not start including a bunch of separate cases.
This, however, has a very unfortunate issue that most (if not all) texts in set theory tend to overlook: if $\alpha^\beta=\sup\{\alpha^\gamma\mid\gamma<\beta\}$ for a limit ordinal $\beta$, then taking $\alpha=0$ and $\beta=\omega$ we get that $0^n$ is either $1$ if $n=0$ or $0$ otherwise, which means that the supremum gives us $1$ rather than $0$.
Of course, this can be corrected by requiring $\sup\{\alpha^{\gamma+1}\mid\gamma<\beta\}$ or $\limsup$ or one of many other variants that we can have that correct for that issue. But the simplest way is to ignore it and let the reader infer this from context—if they even notice the problem.
Asaf Karagila
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Te definition of ordinal exponentiation that I use says that, for limit $\beta$, $\alpha^\beta=\sup{\alpha^\gamma:0<\gamma<\beta}$. I'm pretty sure I learned this from a textbook, and when I teach it I always point out the reason for the $0<\gamma$ part. – Andreas Blass May 26 '18 at 01:41
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@Andreas: Interesting. I cannot recall a book with such definition. Maybe Azriel's book? I can believe that... :) – Asaf Karagila May 26 '18 at 08:54