Arithmetic of uncountable ordinal

Question

Assume $\alpha$ is an ordinal such that $\alpha \geq \omega_1$. Is it true then that $\alpha = \omega + \alpha$ with respect to ordinal arithmetic?

Answer 1

In fact, $\alpha=\omega+\alpha\iff\alpha\ge\omega^2.$

Suppose $\alpha=\omega+\alpha.$ Then, by induction, $\alpha=\omega\cdot n+\alpha\ge\omega\cdot n$ for every $n\lt\omega,$ whence $\alpha\ge\omega^2.$

Suppose $\alpha\ge\omega^2.$ Then $\alpha=\omega^2+\xi$ for some ordinal $\xi.$ Since$$\omega+\omega^2=\omega\cdot1+\omega\cdot\omega=\omega(1+\omega)=\omega\cdot\omega=\omega^2,$$ it follows that $$\omega+\alpha=\omega+\omega^2+\xi=\omega^2+\xi=\alpha.$$