For finite cardinals $A < B$, it's clear that the powerset $2^A$ is smaller than the powerset $2^B$.
However, it seems that when $A < B$ are infinite cardinals, the strict inequality cannot be proven from ZFC: Easton's Theorem tells us that in general, unless something is known about the cofinalities $\text{cf}(A), \text{cf}(B)$, the most we can say is that $2^A \leq 2^B$. (That is, there are models for which both $2^A = 2^B$ and $2^A < 2^B$.)
My question is about the smallest possible case: $A = \aleph_0$, $B = \aleph_1$. If we had that $\aleph_1 = 2^{\aleph_0}$ (Continuum Hypothesis) then obviously we would have $2^{\aleph_0} < 2^{(2^{\aleph_0})} = 2^{\aleph_1}$, but it seems like that's probably a lot stronger than we would need to show $2^{\aleph_0} < 2^{\aleph_1}$. Contrapositively, if $2^{\aleph_0} = 2^{\aleph_1}$, that would imply the negation of the Continuum Hypothesis, but again, this seems like a very strong sufficient condition for $\neg CH$ that wouldn't be necessary.
Can we show for sure that $2^{\aleph_0} < 2^{\aleph_1}$ or $2^{\aleph_0} = 2^{\aleph_1}$? Is there simply no canonical way to answer this (any more than for the Continuum Hypothesis)? What is the status of this question, and are there any well-known open statements to which it is equivalent? I played around with it in a desultory way to discover why approaches like diagonalization or Konig's theorem wouldn't work to establish an injection $2^{\aleph_1} \to 2^{\aleph_0}$ or lack of same, but maybe there are some cleverer techniques out there I'm not familiar with.
Nope, it's consistent with $\mathsf{ZFC}$ that $2^{\aleph_0}=2^{\aleph_1}$. And Easton's theorem already tells us this: for example, the function sending $\kappa$ to $\kappa^+$ for $\kappa>\aleph_0$ and sending $\aleph_0$ to $\aleph_2$ satisfies the requirements in Easton's theorem.
(Of course Easton is overkill here: just start with a model of $\mathsf{CH}$ and add $\aleph_2$-many Cohen reals. This is what Cohen did in his original model of $\mathsf{ZFC+\neg CH}$ if memory serves.)
Amusingly, $2^{\aleph_0}=2^{\aleph_1}$ can happen for non-silly reasons: the proper forcing axiom $\mathsf{PFA}$ proves that $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$, for example.)