When does $\omega \cdot \alpha = \alpha$?

Question

A proof I am looking at mentions the special case that $\omega \cdot \alpha = \alpha$, but for what ordinals $\alpha$ is this the case?

Answer 1

If $\alpha=\omega^\omega\cdot \beta$ for some ordinal $\beta$, then clearly $\omega\cdot \alpha=\alpha$. By considering the Cantor normal form of $\alpha$ (or, even better, by writing $\alpha$ in base-$\omega^\omega$ positional numeral system), we obtain that $\alpha=\omega^\omega\cdot \beta$ is also necessary.