The $\aleph$ hierarchy, I think, is a sequence of infinite cardinals. $\aleph_0$ is $|\mathbb{N}|$, possibly by definition. $\aleph_1$ is the next highest cardinal.
A lot has been written about whether $\aleph_1 = |\mathbb{R}|$ or, equivalently $\aleph_1 = \beth_1$ and how the continuum hypothesis is independent of ZFC.
But there definitely isn't a cardinal $\kappa$ such that $\aleph_0 < \kappa < \aleph_1$.
Why do we know that there's a successor cardinal to $\aleph_0$ / least cardinal greater than $\aleph_0$?
What would break if we insist, by fiat, on the existence of a hierarchy of cardinals indexed by the non-negative reals or non-negative rationals (in the way that $\aleph_n$ is indexed by $n \in \mathbb{N_0}$)? If cardinals aren't primitive things and the existence of a successor falls right out of how they're defined, this question might not make sense.
First: assuming the axiom of choice we have that the class of cardinals is equal to the union of all limit ordinals hence equal to $On$(or $Ord$) thus it is well ordered.
If we do not assume choice we need to go back to the definition of $\aleph_a$, we have $\aleph_0=\omega$, and $\aleph_{a+1}=\aleph_a^+$ and if $a$ is limit ordinal then $\aleph_a=\bigcup\{ \aleph_b\mid b< a\}$, now because ordinals are well founded there exists unique successor cardinal for every aleph number, thus aleph one is well defined