Here is my proof, given that $|A|=a$ and $\omega=|\mathbb{N}|$(also commonly denoted as $\aleph_0$):
Let $F$ be the set of all bijections of the form from $B^{\omega}$ to $B$ where $B\subset A$ ($F$ is not empty, since $(2^{\omega})^{\omega}=2^{\omega}$);
Order $F$ using extension of function, apply Zorn's Lemma for a maximal function $g:B^{\omega}\to B$;
To show that $|B|=|A|$, notice that otherwise $A\setminus B$ contains a set $C$ with the same cardinality as $B$, and so $g$ can be extended to $(B\cup C)^{\omega}\to (B\cup C)$, which violates the maximality of $B$;
Since $|B|=|A|=a$ and $|B^{\omega}|=|B|$, we have $a^{\omega}=a$.
This looks very fishy, as I failed to discover a theorem that directly implies this. Is the proof above correct?
The claim is false in the first place. A consequence of Konig's theorem is that for all infinite cardinals $\kappa$, it holds $$\kappa<\kappa^{\operatorname{cof}\kappa}$$ and there is plenty of cardinals $\kappa$ such that $\kappa>2^{\aleph_0}$ and $\operatorname{cof}\kappa=\aleph_0$. Notably, $\beth_\omega$ (see here for the definition of beth-sequence).