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This question already has an answer here:

How would you prove that $$n^{\gamma} = m^{\gamma}$$ for every limit ordinal $\gamma$ and $n,m$ finite ordinals?

It's a rather short solution problem, but I can't construct any slick answer for it. I know very little about ordinal exponentiation, just that $\alpha^{\beta +1} = \alpha^{\beta}\cdot\alpha$ and that if there's a limit ordinal in the exponent we take the $\sup$.

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marked as duplicate by Asaf Karagila set-theory Mar 22 '16 at 19:35

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    $\begingroup$ That was LITERALLY the first question in the [ordinals] tag at the time when you posted your question. $\endgroup$ – Asaf Karagila Mar 22 '16 at 19:35
  • $\begingroup$ Yeah that's one of the questions in a problem's list in a set theory class at my Uni, seems I'm not the only one having trouble with it ;) $\endgroup$ – Jules Mar 22 '16 at 19:39
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You need to assume that $m,n>1$.

First show that it’s true for $\gamma=\omega$. Then show that if it’s true for some limit $\gamma$, it’s true for $\gamma+\omega$; this is actually a bit like the first step. Finally, show that if it’s true for every limit ordinal less than $\gamma$, and $\gamma$ is a limit of limit ordinals, then it’s true for $\gamma$ as well.

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    $\begingroup$ Yes, you're right, I forgot about this assumption! $\endgroup$ – Jules Mar 22 '16 at 19:33

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