Prove $\omega^{\omega_1}=\omega_1$ and $2^{\omega_1}=\omega_1$ Prove that $\omega^{\omega_1}=\omega_1$ and $2^{\omega_1}=\omega_1$.
I also found an exercise asking to compute the ordinal number $\omega_1^{\omega}$, but I do not even understand what I am supposed to do, any help?
 A: Ordinal exponentiation is not cardinal exponentiation.
Recall the definition of $\alpha^\beta$:


*

*$\alpha^0=1$.

*$\alpha^{\beta+1}=\alpha^\beta\cdot\alpha$.

*$\alpha^\beta=\sup\{\alpha^\gamma\mid\gamma<\beta\}$ for a limit ordinal $\beta$.


So now we note the following:


*

*If $\alpha,\beta$ are countable ordinals then $\alpha^\beta$ is countable. We can prove this by induction.

*If $\beta>0$ then $\alpha<\alpha^\beta$.


From this it is somewhat easy to calculate $2^{\omega_1}=\omega^{\omega_1}$, since for every $\beta<\omega_1$ we have $2^\beta$ and $\omega^\beta$ to be countable ordinals, and this is a strictly increasing sequence of length $\omega_1$.

As for the additional exercise for $\omega_1^\omega$, I'm not clear about what it means "to compute the ordinal number", because $\omega_1^\omega$ is an ordinal number.
If one interprets "the ordinal number" as the Cantor normal form of $\omega_1^\omega$, then one can note that:
$$\omega_1^\omega=\left(\omega^{\omega_1}\right)^\omega=\omega^{\omega_1\cdot\omega}$$
I don't know if that's simpler, though.
