Does $\alpha^{\omega}$ have a countable fixed point $\alpha>1$? In ordinal arithmetic, does the function $f(\alpha)=\alpha^{\omega}$ have a fixed point which is countable and greater than 1? I think the answer is "no", as I've tried quite hard to construct examples, but I do not understand how to use the countability. Really, I would like to prove $\alpha$ countable implies $\alpha < \alpha^{\omega}$ for $\alpha >1$. I can do this via an induction but I don't see how to treat the case where $\alpha$ is a limit.
I would appreciate guidance in how to show that functions defined by arithmetic operations on ordinals have no countable fixed point more generally.

You are correct. There are none. To see why, note that there is a monotonicity here: $$\alpha<\alpha\cdot\alpha=\alpha^2<\dots<\alpha^\omega,$$
for any $\alpha>1$. 
To prove this, simply note that $\alpha+1<\alpha^2$ as well. And therefore $\alpha$ is a proper initial segment of $\alpha^2$, and thus of $\alpha^\omega$.
As a final remark, see that the above doesn't even use the fact that $\alpha$ is countable.
