 Solution

For every countable ordinal $\beta$, ${\omega }^{\beta }$ is countable: successor stages use a countable product, and countable limit stages a countable union. Hence ${\omega }^{{\omega }_1}={\sup }_{\beta <{\omega }_1}{\omega }^{\beta }\le {\omega }_1$. Conversely ${\omega }^{\beta }\ge \beta$ by transfinite induction, so these values are cofinal in ${\omega }_1$. Therefore ${\omega }^{{\omega }_1}={\omega }_1$.

It is not the least fixed point. Set $a_0=1$, $a_{n+1}={\omega }^{a_n}$ and define epsilon zero as ${\epsilon }_0={\sup }_n{a}_n$. This is countable, and continuity of ordinal exponentiation at limits gives ${\omega }^{{\epsilon }_0}={\sup }_n{\omega }^{a_n}={\epsilon }_0$. Any fixed point of $\alpha \mapsto {\omega }^{\alpha }$ dominates every $a_n$ by induction, hence dominates ${\epsilon }_0$. Thus the least fixed point is ${\epsilon }_0<{\omega }_1$.