Set theory problem: arithmetics of cardinal numbers

Question

Let $a$ and $b$ be any cardinal numbers, $a^b$ is equal to cardinal number of the set of natural numbers. Prove that $a$ is equal to cardinal number of the set of natural numbers and $b$ is an element of $\{0,1,2,\ldots\}$. Is it possible to prove it by induction?

Answer 1

Given sets $A$ and $B$. How many function $f:B\to A$ exist (because this is the cardinality of $|A|^{|B|}$? We can check some cases:

1. If $B$ is empty then there is exactly one function. In the following, assume that $B$ is not empty.
2. If $A$ is empty, then there are no functions at all. If $A$ has one element, then there is only a single function. In the following assume that $A$ has more than one element.
3. For every $a\in A$ there is the function $f(b)=a$ for all $b\in B$. Hence there are at least as many such functions as elements in $A$.
4. Let $a_1,a_2\in A$ distinct elements. For every $b^*\in B$ there is the function $$f(b)=\begin{cases}a_1 & \text{if $b=b^*$} \\ a_2 & \text{otherwise} \end{cases}.$$ So there are at least $B$ many functions.
5. If $A$ and $B$ are finite, then there are exactly $|A|^{|B|}$ many function.

These considerations show that neither $A$ nor $B$ can be empty, that $A$ must have at least two elements, that neither $A$ nor $B$ can be uncountable, and that at least one of the sets must be infinite.

So we have the three cases 1. $A$ finite, $B$ countably infinite, 2. $A$ countably infinite, $B$ finite, 3. $A$ and $B$ countably infinite. The following well-known observation excludes two of these options:

$${\aleph_0}^{\aleph_0} = n^{\aleph_0} =2^{\aleph_0}=\mathfrak c>\aleph_0,\qquad\text{for $n\ge 2$}.$$

The only case remaining is that $A$ is countable and $B$ is finite (but not $0$).