Show that if $\kappa$ is an uncountable cardinal, then $\kappa$ is an epsilon number Firstly, I give the definition of the epsilon number: 

$\alpha$ is called an epsilon number iff $\omega^\alpha=\alpha$.

Show that if $\kappa$ is an uncountable cardinal, then $\kappa$ is an epsilon number and there are $\kappa$ epsilon numbers below $\kappa$; In particular, the first epsilon number, called $\in_0$, is countable.
I've tried, however I have not any idea for this. Could anybody help me?

The following is intended as a half-outline/half-solution.
We will prove by induction that every uncountable cardinal $\kappa$ is an $\epsilon$-number, and that the family $E_\kappa = \{ \alpha < \kappa : \omega^\alpha = \alpha \}$ has cardinality $\kappa$.
Suppose that $\kappa$ is an uncountable cardinal such that the two above facts are knows for every uncountable cardinal $\lambda < \kappa$.


*

*If $\kappa$ is a limit cardinal, note that in particular $\kappa$ is a limit of uncountable cardinals.  By normality of ordinal exponentiation it follows that $$\omega^\kappa = \lim_{\lambda < \kappa} \omega^\lambda = \lim_{\lambda < \kappa} \lambda = \kappa,$$ where the limit is taken only over the uncountable cardinals $\lambda < \kappa$.
Also, it follows that $E_\kappa = \bigcup_{\lambda < \kappa} E_\lambda$, and so $| E_\kappa | = \lim_{\lambda < \kappa} | E_\lambda | = \kappa$.

*If $\kappa$ is a successor cardinal, note that $\kappa$ is regular.  Note, also, that every uncountable cardinal is an indecomposable ordinal.  Therefore $\kappa = \omega^\delta$ for some (unique) ordinal $\delta$.  As $\omega^\kappa \geq \kappa$, we know that $\delta \leq \kappa$.  It suffices to show that $\omega^\beta < \kappa$ for all $\beta < \kappa$.  We do this by induction: assume $\beta < \kappa$ is such that $\omega^\gamma < \kappa$ for all $\gamma < \beta$.


*

*If $\beta = \gamma + 1$, note that $\omega^\beta = \omega^\gamma \cdot \omega = \lim_{n < \omega} \omega^\gamma \cdot n$.  By indecomposability it follows that $\omega^\gamma \cdot n < \kappa$ for all $n < \omega$, and by regularity of $\kappa$ we have that $\{ \omega^\gamma \cdot n : n < \omega \}$ is bounded in $\kappa$.

*If $\beta$ is a limit ordinal, then $\omega^\beta = \lim_{\gamma < \beta} \omega^\gamma$.  Note by regularity of $\kappa$ that $\{ \omega^\gamma : \gamma < \beta \}$ must be bounded in $\kappa$.
To show that $E_\kappa$ has cardinality $\kappa$, note that by starting with any ordinal $\alpha < \kappa$ and defining the sequence $\langle \alpha_n \rangle_{n < \omega}$ by $\alpha_0 = \alpha$ and $\alpha_{n+1} = \omega^{\alpha_n}$ we have that $\alpha_\omega = \lim_{n < \omega} \alpha_n < \kappa$ is an $\epsilon$-number.  Use this fact to construct a strictly increasing $\kappa$-sequence of $\epsilon$-numbers less than $\kappa$.
(There must be an easier way, but I cannot think of it.)
