Cardinality Of $n^{\aleph_0}$ for cardinals $n$ up to $\aleph_0$? Considering The Continuum Hypothesis in it's most simplified form States that $|\mathbb{R}| = 2^{\aleph_0} = \aleph_1$.

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*Is $3^{\aleph_0} = |\mathbb{R}| = 2^{\aleph_0} = \aleph_1$?
what about other $n\in N: n^{\aleph_0}?$ For $n=1$, it looks like $\aleph_0$ to me :) so what would it be for $n=0$ ?!



*What if $|\mathbb{R}| = \aleph_0^{\aleph_0} = \aleph_{\aleph_0} > ... > 4^{\aleph_0} > 3^{\aleph_0} > 2^{\aleph_0} > 1^{\aleph_0} = \aleph_0$, there were countably infinite many continuums, and call $\mathbb{R}$ something like The Absolute Continuum?

PS: just learned a bit of latex to write this post plz don't judge my language. i'm more used to turing-complete universes :)
send bacon and huggies
H
 A: In this context, the notation $a^b$ is the size of the set of functions from a set of size $b$ to one of size $a$.  For example, you can check for yourself that there are exactly 8 possible functions from a set of size 3 to one of size 2.
The notation $n^{\aleph _0}$ means the number of dfferent fuctions from $\Bbb N$ to a set with $n$ elements.  We can take our set of  $n$ elements to be $\{0,1,2,\ldots ,n-1\}$.   And we can understand a function from $\Bbb N$ to $\{0,1,2,\ldots ,n-1\}$ as an infinite sequence of elements of $\{0,1,2,\ldots ,n-1\}$. ( In fact, this is the formal definition of “sequence”.)
It can be shown that for $n≥2$ the set of sequences is always the same size.  Here is one unusually straightforward   example. Suppose $b_1, b_2, b_3, b_4,\ldots$ is a sequence whose elements are all $0$’s and $1$’s. Then
$$b_1\cdot 2+b_2,
b_3\cdot2+b_4, \ldots$$ is a sequence whose elements are all from $\{0,1,2,3\}$, and each sequence of elements of $\{0,1,2,3\}$ similarly corresponds to exactly one sequence of elements of $\{0,1\}$.
For other $n≥2$ the situation is the same. Note that each real number between 0 and 1 usually corresponds to a single sequence of elements of $\{0,1,\ldots n-1\}$: its representation as a sequence of base-$n$ digits. (There are some exceptions, but only a few, and these don't affect the final count.)
For $n=1$ the situation is simpler. There is only one sequence:
$$1,1,1,1,1,\ldots$$ so $$1^{\aleph_0}=1.$$
For $n=0$, there are no sequences because the set of permitted elements is empty: $$0^{\aleph_0}=0.$$
It can be shown that $\aleph_0^{\aleph_0} =2^{\aleph_0}$. But $\aleph_{\aleph_0}$ (usually written $\aleph_\omega$ for technical reasons) is much, much larger, far larger than $\aleph_1$.
A: for qn 1, $1^{\aleph_0}=1$ and $0^{\aleph_0}=0$
for qn 2,
assume $n\geq 2$ is a natural number.
$2\leq n \leq \aleph_0$
$2^{\aleph_0}\leq n^{\aleph_0}\leq \aleph_0^{\aleph_0}\leq (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0}$
so every element in the ineq chain is $2^{\aleph_0}$.
i don't know what $\aleph_{\aleph_0}$ is though, but it is larger than ${\aleph_1}$.
the above argument shows $2^{\aleph_0}=4^{\aleph_0}$ , but one can prove this in another way.
$4^{\aleph_0}=2^{2\aleph_0}=2^{\aleph_0}$
