I am not a mathematician, but rather a computer engineer with a curious mind.
The continuum hypothesis states (I believe) that there does not exist a set $S$ such that $\aleph_0 < |S| < 2^{\aleph_0}$
The cardinality $2^{\aleph_0}$ always seems to be taken for granted and not really defined in explanations that I run across.
My interpretation of this value is the following:
$2^{\aleph_0} = $ the number of ways of choosing between 2 possibilities at least 0 times and at most $\aleph_0$ number of times.
Is this an appropriate interpretation?
The main reason I thought this even might be true: if this is true, then $2^{\aleph_0}$ is like the cardinality of the binary representations of real numbers, whereas something like $10^{\aleph_0}$ would be the cardinality of the decimal representations of of real numbers. This led me to believe $n^{\aleph_0}$ are equal for ALL $n \in \mathbb{N}, n > 1$. This also showed to me that unary numbers cannot represent the real numbers (because my definition would lead to $1^{\aleph_0} = \aleph_0$) which I found interesting and (at least intuitively) consistent.
$2^{\aleph_0}$ is defined to be the number of ways of deciding between two alternatives exactly $\aleph_0$ times.
This can be restated as: $2^{\aleph_0}$ is the number of infinite sequences of 0s and 1s, where the sequence has one entry for each natural numbers.
So you are correct that $n^{\aleph_0}$ is, when $n \geq 2$, the number of $n$-ary expansions of numbers in the interval $[0,1]$.
The number of ways of deciding between two alternatives a finite number $m$ times is $2^m$. We have $\bigcup_{m \geq 2} 2^m = \aleph_0$, so in your characterization using "at most $\aleph_0$ number of times", the real issue is with the sequences the make a choice exactly $\aleph_0$ times.
Regarding the equality at the end of the question, something stronger is true. We can show in ZFC that, for $n \geq 2$, $n^{\aleph_0} = 2^{\aleph_0} = \aleph_0^{\aleph_0}$.