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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

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    See https://en.wikipedia.org/wiki/Continuum_hypothesis#Implications_of_GCH_for_cardinal_exponentiation for some facts, that might be interesting for you. – 2017-01-28

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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.

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    Sorry for questioning, but did you mean "with or without continuum hypothesis"? I thought Koenig's lemma directly implied the axiom of choice as a corollary. – 2017-01-28
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    There is an obvious injection from $\aleph_\omega$ to $\aleph_\omega^{\aleph_0}$. If the latter can be well-ordered, then Koenig's lemma applies; if it cannot be well-ordered, then the cardinalities cannot be equal anyway. – 2017-01-28
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    Oh... ZF is a place of sneaky tricks. $\ddot\smile$ – 2017-01-28