1.11.1. Prove that ${\omega}^2=\omega \omega$ is not of the form $\delta + \omega$ for some ordinal $\delta$.
This is one of the exercises in the book "A First Journey through Logic". I tried to prove it, but I am not sure about that. If someone give me a feedback about the proof, it would be really helpful.
Proof. Suppose not. Since $\omega$ is a limit ordinal, by the properties of ordinal addition and multiplication, we obtain $\bigcup_{\beta < \omega }\delta + \beta =\delta + \omega = \omega \omega= \bigcup_{\alpha < \omega}\omega \alpha$ .
Now, taking $\beta =0$, we get $\delta < \omega \alpha $ for some $\alpha < \omega $ . Since $\omega \alpha $ is a limit ordinal (this is the first part of the exercise), we get $\delta + \gamma < \omega \alpha $ for all $\gamma <\omega$. Since $\alpha<\omega$ implies $\omega\alpha<\omega \omega$, we conclude that $ \delta + \omega \leq \omega \alpha < \omega \omega = \delta + \omega $. This contradiction finishes the proof.
P.S. This is my first question in this website. If you have any suggestion about my writing, please let me know.