# Relation between $\omega^{\omega_1}$ and $\omega_1^{\omega}$

If $x,y$ are two ordinals, then we have three possible cases: $x=y$, $x\in y, y \in x$ (here $<$ also means $\in$). I would like to know which case happen for $x=\omega^{\omega_1}$ and $y=\omega_1^{\omega}$, where I consider the ordinal operations.

With these questions, we know that $x=\omega_1$. Then $y$ is the union of $\omega_1^n$ for $n<\omega$, but I don't even know how to compute $\omega_1^n$. For instance, $\omega_1^2$ is the union of $\omega_1\cdot \gamma$ for $\gamma < \omega_1$. I don't think that $\omega_1^2$ can be simplified.

As you say, $\omega^{\omega_1}=\omega_1$; this is because $\omega^\alpha\ge \alpha$ for any ordinal $\alpha$, and is countable for any countable ordinal $\alpha$.
Meanwhile, you're right that $\omega_1^n$, and $\omega_1^\omega$, can't be simplified. However, you don't need to simplify them in order to compare them with $\omega_1$! Just note that $$\omega_1<\omega_1+1\le \omega_1^2\le\omega_1^\omega.$$ (In fact, both the "$\le$"s are strict, but that's not necessary here.) This is enough to show that $x<y$.