= Solution
For every countable <ordinal> $\beta$, $\omega^\beta$ is countable: successor stages use a countable product, and countable limit stages a countable union. Hence $\omega^{\omega_1}=\sup_{\beta<\omega_1}\omega^\beta\leq\omega_1$. Conversely $\omega^\beta\geq\beta$ by <transfinite induction>, so these values are cofinal in $\omega_1$. Therefore \b[$\omega^{\omega_1}=\omega_1$].
It is not the least fixed point. Set $a_0=1$, $a_{n+1}=\omega^{a_n}$ and define <epsilon zero> as $\varepsilon_0=\sup_na_n$. This is countable, and continuity of <ordinal exponentiation> at limits gives $\omega^{\varepsilon_0}=\sup_n\omega^{a_n}=\varepsilon_0$. Any fixed point of $\alpha\mapsto\omega^\alpha$ dominates every $a_n$ by induction, hence dominates $\varepsilon_0$. Thus \b[the least fixed point is $\varepsilon_0<\omega_1$].
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