Cardinal arithmetic does not seem to open its way to the existence of $\aleph_1$ that is not $2^{\aleph_0}$, as any operation on $\aleph_0$ would lead to $\aleph_0$ or $2^{\aleph_0}$ and $2^{2^{\aleph_0}}$ or so forth.
According to my knowledge, choice does not determine whether continuum hypothesis is right or wrong, so what is going on with this?
No, cardinal arithmetic is equipped with a successor function, much like the natural numbers. The successor of a cardinal $\kappa$ is the smallest cardinal which is strictly larger than $\kappa$. The successor is denoted by $\kappa^+$.
$\aleph_1$ exists (regardless to the axiom of choice) because we can prove there exists an uncountable ordinals, and therefore there exists a least uncountable ordinal. By definition of being the least uncountable ordinals it is a cardinal (=an ordinal which is not in bijection with any smaller ordinal), and it is exactly $\aleph_1$.
You are correct however that the axiom of choice does not determine the value of $2^{\aleph_0}$, it is however needed to ensure that it is a well-ordered cardinal.
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