I need some help with some basic ordinal arithmetic. I am trying to determine the supremum of the sequence $\omega^{2}*1 + \omega$, $\omega^{2}*2 + \omega$, $\omega^{2}*3 + \omega$, ... where the sequence $1$, $2$, $3$, ... is the natural number sequence.
I think that the answer is $\omega^{3}$. My reasoning is a little clumsy, though. Suppose we replaced the $+$ $\omega$ term with $+$ $\omega^{2}$ instead. Then, the sequence would simply be: $\omega^{2}*2$, $\omega^{2}*3$, $\omega^{2}*4$, ..., which obviously has supremum $\omega^{3}$. Now, the original sequence clearly can't be any greater than this sequence, and since it also involves $\omega$ many $\omega^{2}$ terms, it must be at least $\omega^{3}$. Thus, the answer is $\omega^{3}$.
(1) Is this answer correct? (2) What would a more elegant proof be?
I think that your idea is as elegant as it gets, but personally I'd rather be more efficient with my theorems.
Ordinal operations $+$ and $*$ are strictly increasing in the second variable and weakly increasing in the first variable (and, for the record, so is exponentiation as long as the base is $\ge2$). Therefore for all $n<\omega$ we have $$\omega^2* n+\omega< \omega^2*n+\omega^2=\omega^2*(n+1)<\omega^3$$
Therefore $\sup_{n<\omega} (\omega^2*n+\omega)\le \omega^3$.
Moreover, $\omega^2*n+\omega>\omega^2*n$, therefore $\sup_{n<\omega}(\omega^2*n+\omega)\ge \sup_{n<\omega}\omega^2*n=\omega^3$.