Why can we choose a greatest ordinal $\beta$, such that $\omega^\beta\leq \alpha$?

Question

I am reading a proof of Cantor's normal form theorem. In it, I read:

for arbitrary $\alpha>0$ let $\beta $ be the greatest ordinal such that $\omega^\beta \leq \alpha$.

Why should such an ordinal exist?

Answer 1

There are ordinals $\varepsilon$ such that $\omega^{\varepsilon} > \alpha$. By the well-ordering of the ordinals, there is hence a smallest ordinal $\gamma$ with $\omega^{\gamma} > \alpha$.

For limit ordinals $\lambda$, we have

$$\omega^{\lambda} = \bigcup_{\delta\in \lambda} \omega^{\delta} = \sup \{ \omega^{\delta} : \delta \in \lambda\}.$$

Since $\omega^{\delta} \leqslant \alpha$ for all $\delta \in \gamma$ by definition of $\gamma$, we have

$$\sup \{ \omega^{\delta} : \delta \in \gamma\} \leqslant \alpha < \omega^{\gamma},$$

hence $\gamma$ is not a limit ordinal, and thus there is a $\beta$ with $\gamma = \beta + 1$. This is then the largest ordinal with $\omega^{\delta} \leqslant \alpha$.