Prove: If $\alpha > 1$, then $\omega\alpha > \omega$. If at the same time $\alpha < \omega$, we have $\alpha\omega = \omega$.
The first part is more or less straight forward by transfinite induction with base case starting from 2.
I'm having more trouble with the second part. It seems reasonable to do induction as well.
Base case: $\alpha=2$, then $\alpha\omega = type(2\times\omega, L)$ where L is the lexicographical order. I can define a mapping $f: 2\times\omega\rightarrow\omega$ to prove they are indeed equal. However, I'm having quite some trouble dealing with induction step.
If $\alpha\omega = \alpha$ for some $1 < \alpha < \omega$. Then, how would I prove $S(\alpha)\omega = (\alpha + 1)\omega = \omega$? I don't see how I might use the induction hypothesis to ease the proof here.