# Show that for $n, m \in \omega$, the ordinal and cardinal exponentiations $n^m$

This is an exercise from Kunen's book.

Show that for $n, m \in \omega$, the ordinal and cardinal exponentiations $n^m$ are equal.

What I've tried: I want to prove by using induction on $m$. For $m=0$, the ordinal exponentiation $n^m=n^0=1$ by the definition; and the cardinal exponentiation $n^m=n^0=|n^0|=1$. Now we assume for $m=k$ the case is right. Then for the $m=k+1$, ordinal exponentiation $n^{k+1}=n^k\times n$ and the cardinal exponentiation $n^{k+1}=|n^{k+1}|=|n^k\times n|$. Then I don't how to go on.

Could anybody help me? Thanks ahead:)

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## 1 Answer

First show that addition and multiplication also behave the same with ordinal and cardinal exponentiation when restricting them to natural numbers.

Start with addition, that should be quite easy. Then use this to show that multiplication also has this property. From there you can use this to continue from where you got stuck.

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Thanks Asaf. Let me try. I will come back to you if anything I cannot pass. –  Paul Aug 9 '12 at 11:53