Find ordinals $\alpha$ such that (a) $n^{\alpha}=\alpha\; $ (b) $\omega_1^{\alpha}=\alpha$
On (a) I could verify that all ordinals of the form $\omega, \omega^{\omega},\omega^{\omega^\omega},\cdots$ satisfy the equation, but are they all of them? Does it work for $\alpha=\omega_1$? For all $\omega_{\alpha}$? Well, $n^{\omega_1}=\sup_{\delta<\omega_1} n^{\delta}$ is this equal to $\omega_1 ?$ I know certainly that it is $\geq$ but couldn't prove $\leq$. And if it does holds I think I could repeat this argument for all ordinals $\omega_{\alpha}$.
On (b) I couldn't get anything. Does it holds for all cardinals? I thought following the same argument on (a) but it doesn't seem to be very trustful.
Could you help me?
Let $\epsilon_\alpha$ be the $\alpha$th solution of the equation $\alpha=\omega^\alpha$. They are known as epsilon numbers. Then we have
$\alpha=\omega$ is the simplest solution. Moreover, $\omega$ is the only possible solution that are less than $\omega^2$. (Just put $\alpha=\omega\cdot k+l$ into the equation $n^\alpha=\alpha$.) Now assume that $\alpha\ge\omega^2$. I claim that $\alpha=\epsilon_\beta$ for some $\beta$.
If $n\ge 2$, then $n^{\epsilon_\beta}=\epsilon_\beta$. This follows from some simple ordinal inequalities.
On the other hand, we can see that $n^\alpha=\alpha$ implies $\omega^\alpha=\alpha$: $\alpha\ge\omega^2$ implies $\alpha=\omega+\alpha$. From this, we can show that $\omega\cdot\alpha=n^\omega n^\alpha=n^{\omega+\alpha}=\alpha$ and $\omega^\alpha = (n^\omega)^\alpha = n^\alpha=\alpha$. (I use the equality $n^\omega=\omega$.)
(The previous solution does not consider the case $\alpha<\omega^2$. Thanks to @Simply Beautiful Art for pointing it out.)
Clearly, every ordinal which satisfies $\omega_1^\beta=\beta$ is an epsilon number. However, not every epsilon number satisfies $\beta=\omega_1^\beta$: You can see that $\omega_1$ is an epsilon number, and $\omega_1^{\omega_1}>\omega_1$.
However, if $\beta$ is an epsilon number greater than $\omega_1^\omega$, then we have $\omega_1^\beta=(2^{\omega_1})^\beta = 2^\beta\le \beta$. @bof has already shown that if $\omega_1^\beta=\beta$, then $\beta>\omega_1^\omega$, so we have all possible solutions.