Is there an ordinal number $x$ such that $\omega^\omega \cdot x =\omega_1$?
My work: After googling, I found this which states the lemma below:
Lemma. If $\square$ is any ordinal arithmetic operation, and at least one of $\alpha,\beta$ is infinite, then $|\alpha\square\beta|=\max\{|\alpha|,|\beta|\}$.
I assumed $||$ here means cardinality. So according to this, if $x$ satisfies $\omega^\omega \cdot x =\omega_1$, then we have $\max\{|\omega^\omega|,|x|\}=|\omega_1|$, and since $|\omega^\omega|$ is countable, $|x|$ must be uncountable.
So I thought substituting $x=\omega_1$ would be a nice start, but I don't know how to calculate $\omega^\omega \cdot \omega_1$.
How can I proceed from here?
By the definition of ordinal multiplication, since $\omega_1$ is a limit ordinal, $\omega^\omega\cdot\omega_1$ is the supremum of all products $$ \omega^\omega\cdot\gamma $$ for all ordinals $\gamma<\omega_1$. All of these are countable. Which is to say, we know that $\omega^\omega\cdot\omega_1$ is uncountable, and that it is the supremum of a set of countable ordinals. There is only one such ordinal.