Prove that $\omega + \omega_1 = \omega \cdot \omega_1 = \omega^{\omega_1} = \omega_1$

I am assuming already that a) the union of countably many countable sets is countable and b) $\omega_1$ is the least uncountable ordinal, so $x < \omega_1$ if and only if $x$ is a countable ordinal.

I'm not sure if it is relevant but this question also allows The Axiom of Choice.

Thanks

Solution 1:

HINT: Recall the induction definition of all ordinal arithmetic. If $\delta$ is a limit ordinal, then $\alpha\square\delta=\sup\{\alpha\square\gamma\mid\gamma<\delta\}$ (where $\square$ can be addition, multiplication or exponentiation). Now prove the following lemma, and you're about done:

Lemma. If $\square$ is any ordinal arithmetic operation, and at least one of $\alpha,\beta$ is infinite, then $|\alpha\square\beta|=\max\{|\alpha|,|\beta|\}$.