How many epsilon numbers $<\omega_1$ are there? An epsilon number is an ordinal $\epsilon$ such that $\epsilon=\omega^\epsilon.$ What is the cardinality of the set of all epsilon numbers less than $\omega_1$? 
I'm asking this because of a proof I've just read that seems to presuppose that there are countably many such ordinals, and it seems to me intuitively that there should be uncountably many (although I don't know how to prove it).
Added. OK, I've just understood that the proof I mentioned is OK even if there are uncountably many such ordinals, but I still don't see how I can find their number.
 A: Note that $\varepsilon_0$ is countable. So if there are only countably many countable $\varepsilon$ numbers, they would have a countable supremum, $\alpha$. Consider now the same construction as $\varepsilon_0$, starting $\alpha+1$. That is:
$$\sup\{\omega^{\alpha+1},\omega^{\omega^{\alpha+1}},\ldots\}$$
The result is itself an $\varepsilon$ number, and it is countable (as the countable limit of countable ordinals). But all the countable $\varepsilon$ numbers were assumed to be below $\alpha$, which is a contradiction.
A: Your intuition is correct: $\omega_1=\epsilon_{\omega_1}$. See the discussion in Wikipedia: the constructions of $\epsilon_{\alpha+1}$ from $\epsilon_\alpha$ and of $\epsilon_\alpha$ for a countable limit ordinal $\alpha$ preserve countability, so there must be $\omega_1$ countable $\epsilon$-numbers.         
A: Another answer that uses choice:
Since $\varepsilon_\alpha$ is countable if $\alpha$ is countable by this other SE question or by the arguments in the other answers (and obviously $\varepsilon_\alpha\ge\alpha$), the map $f:\beta\to\omega_1$ defined by $f(\alpha)=\varepsilon_\alpha$ (where $\beta$ is the number of countable $\varepsilon$ numbers) is a cofinal map. But by the axiom of choice, $\omega_1$ is regular. Thus there are $\beta=\omega_1$ countable $\varepsilon$ numbers.
