Prove that for $\alpha$ an ordinal, $\alpha \le \omega^\alpha$.

Question

Prove that for $\alpha$ an ordinal, $\alpha \le \omega^\alpha$.

I assume the easiest proof is via induction, but I do not know what to do if $\alpha$ is a limit ordinal.

Answer 1

The proof by induction is indeed the easiest.

If $\alpha$ is limit, then $\omega^\alpha=\sup\{\omega^\beta\mid\beta<\alpha\}$. Now use the induction hypothesis, and the following hint:

If $A,B$ are two sets of ordinals, and for every $\alpha\in A$ there is $\beta\in B$ such that $\alpha\leq\beta$. What is the relationship between $\sup A$ and $\sup B$?