Cardinality Of $n^{\aleph_0}$ for cardinals $n$ up to $\aleph_0$?

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 $\aleph_1$.