# 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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