What is the result of a number greater than 2 raised to the power of {Aleph-0}? So, I know that $2^{\aleph_0} = \beth_1$, right? What about another number, say $10$, raised to the power of $\aleph_0$? Is $10^{\aleph_0} = \beth_1$ also true, or is $10^{\aleph_0} > \beth_1$ somehow?
 A: We have $2^{\aleph_0} \leq 3^{\aleph_0} \leq 4^{\aleph_0} \leq \dots \leq \aleph_0^{\aleph_0}$, because of the inclusions $\{0, 1\}^\Bbb N \subset \{0,1,2\}^\Bbb N \subset \dots \subset \Bbb N^\Bbb N$. So if we prove that $\aleph_0^{\aleph_0} \leq 2^{\aleph_0}$, then we see that all of these cardinalities are in fact equal. 
To show this, we need to find some injection $f: \Bbb N^{\Bbb N} \to \{0, 1\}^\Bbb N$. There are many ways to do this; my favorite is as follows. Let $a = (a_n)$ be some sequence of natural numbers. Then we define $f(a)$ to be the sequence consisting of first $a_0$ ones, followed by a zero, then $a_1$ ones, followed by a zero, then $a_2$ ones, followed by a zero, and so on. This gives a sequence of zeroes and ones, and if $b = (b_n)$ is another sequence of natural numbers, then $f(a) = f(b)$ if and only if $a_n =b_n$ for all indices $n$ if and only if $a = b$. So $f$ is indeed injective, and therefore $\aleph_0^{\aleph_0} \leq 2^{\aleph_0}$.
So indeed $2^{\aleph_0} = 3^{\aleph_0} = \dots = \aleph_0^{\aleph_0}$. 
A: Any cardinal $\kappa \leq \aleph_{0}$, namely the finite ones and $\aleph_{0}$ itself, gives $\kappa^{\aleph_{0}} = 2^{\aleph_{0}}$. Any cardinal larger than $\aleph_{0}$ remains constant when exponentiated
