Is $\aleph_{1}^{\aleph_0}<\aleph_{2}^{\aleph_0}$ equivalent to the Continuum Hypothesis?

The key is to observe that $\aleph_1^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}$ regardless of whether CH holds. In particular, it's going to help in a bit to not simplify the right hand side.

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Here's a proof of that observation:

• $\le$: immediate, since we know without assumptions that $\aleph_1\le 2^{\aleph_0}$ and raising both sides to $\aleph_0$ doesn't break $\le$.

• $\ge$: the right hand side simplifies to $2^{\aleph_0}$, and (by the same reasoning as the previous bullet) we know that $2^{\aleph_0}\le \aleph_1^{\aleph_0}$.

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Now back to the main problem.

The comparison we're really trying to make is, by the observation above, $$(2^{\aleph_0})^{\aleph_0}\mbox{ versus }(\aleph_2)^{\aleph_0}.$$

We know that we get "$<$" if CH holds. Now suppose CH fails; what does that tell us about $2^{\aleph_0}$ versus $\aleph_2$ (and hence, what does that tell us about the comparison we want to make)?

Using Hausdorff formula we have $$\aleph_2^{\aleph_0}=\aleph_1^{\aleph_0}\cdot\aleph_2,$$ Now using the fact that $\aleph_0^{\aleph_0}\leq\aleph_1^{\aleph_0}\leq (2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$, we get that $$\aleph_2^{\aleph_0}=2^{\aleph_0}\cdot\aleph_2.$$

If $2^{\aleph_0}\geq\aleph_2$, then $\aleph_1^{\aleph_0}=\aleph_2^{\aleph_0}$. Therefore, the assumption that $\aleph_1^{\aleph_0}<\aleph_2^{\aleph_0}$ implies that CH holds.