Given cardinal numbers $A$ and $B$. Does there hold
$$A>B \quad\Rightarrow\quad 2^A>2^B\quad?$$
In other words: is the cardinality of a set uniquely determined by the cardinality of its power set? My first guess was Yes. But then I thought about the continuum hypothesis and that without it there might be a cadinal $\aleph_1$ with $\aleph_0<\aleph_1<\mathfrak c$. I was not sure if $2^{\aleph_1}$ falls nicely between $\mathfrak c$ and $2^{\mathfrak c}$. So it might depend on CH? I have not much experience with cardinal arithmetic, so no clue how to think about this.
At "small" cardinals this can indeed fail when CH does not hold, so (in a model where $\text{MA}(\omega_2)$ holds for example), we have that $2^{\aleph_0} = 2^{\aleph_1} = \aleph_2$, while of course $\aleph_0 < \aleph_1$. So consistently it can fail right away, but if CH holds, then $2^{\aleph_0} = \aleph_1 < 2^{\aleph_1}$, and GCH (the generalised continuum hypethesis) will imply that we have $2^{\aleph_\alpha} = \aleph_{\alpha+1} < 2^{\aleph_{\alpha+1}}$, for all $\alpha$ so that it certainly holds for all cardinals $< \aleph_\omega$ in that case.
In fact, Easton's theorem gives us exactly what the only restrictions on the function $f:\kappa \to 2^\kappa$ are for cardinals $\kappa$. Consistently, we can have a lot of weirdness, but $f$ is always increasing, but not necesarily strictly increasing. E.g. even with GCH we can have $2^{\aleph_\omega} = \aleph_{\omega+2} = 2^{\aleph_{\omega+1}}$. This is a hard theorem to prove (but in Jech's set theory or some other books as well). The singular cardinals hypothesis is also relevant to the cardinality of the power set problem.