Prove that $\alpha + \beta=\beta \iff \alpha\omega\le\beta$
$(\Leftarrow)$ If $\alpha\omega=\beta$, then $\alpha+\beta=\alpha+\alpha\omega=\alpha(1+\omega)=\alpha\omega=\beta.$
If $\alpha\omega<\beta$, then there exists $\gamma$ such that $\beta=\alpha\omega+\gamma$. $\alpha + \beta = \alpha + \alpha\omega+\gamma=\alpha(1+\omega)+\gamma=\alpha\omega+\gamma=\beta$.
$(\Rightarrow)$ Lemma: if $\eta>\xi$ and $k\in\mathbb{N}$, then $\omega^{\xi}k+\omega^{\eta}=\omega^{\eta}$. Indeed, $\eta>\xi \Rightarrow \eta=\xi+\gamma$ $\Rightarrow$ $\omega^{\xi}k+\omega^{\eta}= \omega^{\xi}k+\omega^{\xi+\gamma}= \omega^{\xi}(k+\omega^{\gamma})=\omega^{\xi}\omega^{\gamma}=\omega^{\xi+\gamma}=\omega^{\eta}$.
Let's write $\alpha$ and $\beta$ in Cantor normal form:
$$\alpha=\omega^{\phi_0}a_0+\omega^{\phi_1}a_1+\dots+\omega^{\phi_n}a_n,$$$$\beta=\omega^{\phi_0}b_0+\omega^{\phi_1}b_1+\dots+\omega^{\phi_n}b_n.$$ By lemma, $\alpha+\beta=\omega^{\phi_0}a_0+\omega^{\phi_0}b_0+\omega^{\phi_1}b_1+\dots+\omega^{\phi_n}b_n$. So, for $\alpha+\beta$ to be equal $\beta$, we need $a_0=0$. Now, let's multiply $\alpha=\omega^{\phi_1}a_1+\dots+\omega^{\phi_n}a_n$ by $\omega$, i.e. let's add $\alpha=\omega^{\phi_1}a_1+\dots+\omega^{\phi_n}a_n$ together $\omega$ many times. By lemma, all 'segments' $$\omega^{\phi_2}a_2+\dots+\omega^{\phi_n}a_n+\omega^{\phi_1}a_1$$ would turn into $\omega^{\phi_1}a_1$. Since there are omega many segments of such type, the greatest term of $\alpha\omega$ is $\omega^{\phi_1}a_1\omega=\omega^{\phi_1}(a_1\omega)=\omega^{\phi_1}(\omega)=\omega^{\phi_1+1}\le\omega^{\phi_0}b_0\in\beta.$
I'd like to know whether my proof is correct or not, and if it is, whether it's rigourous enough. It would also be great to see alternative ways of proving this.
Thank you in advance.
For $\Rightarrow$, I'd argue more directly (without Cantor normal form):
Assume $\alpha+\beta=\beta$. Using $\alpha(n+1)+\beta=\alpha n+\alpha+\beta=\alpha n+\beta$, we see by induction that $ \alpha n+\beta=\beta$ for all $n<\omega$. As $\alpha\omega$ is the least ordinal that is $\ge \alpha n$ for all $n<\omega$ and as $\alpha n\le\alpha n+\beta=\beta$, we conclude $\alpha\omega\le\beta$.