Prove that ${\omega}^\beta \le \alpha$

Question

I am completely stuck on proving the lemma.

If $\alpha$ is a non-zero ordinal, then there is a largest ordinal $\beta$ such that ${\omega}^\beta \le \alpha$ and $\beta \le \alpha$

I intuitively see that the first statement is true just from the ordinal arithmetic, but I do not see how $\beta \le \alpha$

Answer 1

Note that $\alpha\leq\omega^\alpha$ by an easy induction argument. So you immediately get the inequality.