I'm trying to prove that if $\omega\underline{\in}\alpha$ and $\alpha^2\underline{\in}\beta$ then $\alpha+\beta=\beta$ where $\alpha,\beta$ are ordinals. I am using transfinite induction for ordinals and inducting on $\beta$. I have already proved $\omega\underline{\in}\alpha$ implies $1+\alpha=\alpha$ (*) so this can be assumed.
I can prove the result assuming $\beta$ is a successor ordinal but I am having difficulty assuming $\beta$ is a limit ordinal. My induction hypothesis is $\zeta\in\beta\rightarrow(\omega\underline{\in}\alpha\land\alpha^2\underline{\in}\zeta\rightarrow\alpha+\zeta=\zeta).$ I am assuming $\omega\underline{\in}\alpha$ and $\alpha^2\underline{\in}\beta$.
So far for the limit ordinal case (which does notactually use the fact that $\beta$ is a limit ordinal) I have proceeded as follows.
As $\alpha^2\underline{\in}\beta$ we know ${\alpha+\alpha}^2\underline{\in}\alpha+\beta$ and so, $\alpha\cdot(\phi^++\alpha)\underline{\in}\alpha+\beta$ and so by *, $\alpha^2\underline{\in}\alpha+\beta $ (**).
Now assume $\alpha+\beta\in\beta$. Then using ** and the induction hypothesis, $\alpha+(\alpha+\beta)=\alpha+\beta $ and so by left cancellation, $\alpha+\beta=\beta$, a contradiction.
Now assume $\beta\in\alpha+\beta$. Then as $\alpha^2\underline{\in}\beta$, either $\alpha^2=\beta$ or $\alpha^2\in\beta$. If $ \alpha^2=\beta$ then $\alpha+\alpha^2=\alpha+\beta$ and so $\alpha\cdot(\phi^++\alpha)=\alpha+\beta$ or, using *, $\alpha^2=\alpha+\beta$ so $\beta\in\alpha^2=\beta$, a contradiction.
The only other case is if $\beta\in\alpha+\beta$ and $\alpha^2\in\beta$, from which I cannot generate a contradiction. If anyone can help, much appreciated. Alternatively, if there is a better approach to the problem, or if I have made any mistakes, that will help as well!
In this scenario it might be more clear how to prove the stronger-looking claim you get by only assuming $\alpha\omega\underline\in\beta$. Since ordinal multiplication is monotonic in the right argument (lemma 3.14 in Schlöder's set of notes "Ordinal Arithmetic"), we have $\alpha\omega\underline\in\alpha\alpha$, so this modified version will imply your original question. This is an attempted proof without transfinite induction.
Assume $\omega\underline\in\alpha$ and $\alpha\omega\underline\in\beta$. Then there must be some $\eta$ such that $\alpha\omega+\eta=\beta$, this isn't supposed to be an application of Euclidean division, instead $\eta$ is supposed to be like "$-\alpha\omega+\beta$" and is well-defined because "Ordinal Subtraction when Possible is Unique". Now we've decomposed $\beta$ into $\alpha\omega$ and an extra part $\eta$, the $\eta$ doesn't matter much other than to get the extra part after the initial $\alpha\omega$ out of the way. Then, $\alpha\omega+\eta=\alpha(1+\omega)+\eta$, which equals $\alpha+\alpha\omega+\eta$ by left distributivity, which then is $\alpha+(\alpha\omega+\eta)$ by associativity of addition, and this whole thing is equal to $\alpha+\beta$ which is what we wanted. In other words, we appended another copy of $\alpha$ to the left of the $\alpha\omega$ part and the ordering still remained isomorphic to $\beta$.