With the axiom of choice it follows that for any infinite cardinal $X$: $$X^n=X$$ where $n\in\mathbb N$.
I'm curious whether or not $X^Y=X$ also holds whenever $Y<X$ and $X$ is infinite. If not, is the classification of those for which it holds a standard result?
The answer is negative, with or without choice.
We can prove that $\aleph_\omega^{\aleph_0}>\aleph_\omega$. Always. This follows from Koenig's lemma.
On the other hand, we cannot even prove that $2^{\aleph_0}=\aleph_1$, so if that fails, we get that $\aleph_1^{\aleph_0}>\aleph_1$. In general, the behavior of the continuum function is just terrible without additional assumptions.
I recommend "Introduction to Cardinal Arithmetic" for an extensive review of basics of cardinal arithmetic, including exponentiation.