Compare the cardinalities of $2^{(2^\mathbb N)}$ and $(2^\mathbb N)^{\mathbb N}$

Question

A set $X^Y$ is defined as the set of all functions from $Y$ to $X$. (Sorry for the mistake earlier where I had written $X$ to $Y$ earlier).

$2^X$ is defined as the set of all subsets of $X$ or equally well as the set of all functions from $X$ to $[2]= 1,2$ .

Please provide a strict inequality/equality w.r.t the cardinalities of the above mentioned sets.

P.S. I am not sure which of the tags elementary-set-theory or set-theory is more suitable for this question.

EDIT: A slight clarification. The only thing I know about cardinalities is that there are five relations defined on them, the equality and the four inequalities, with their standard properties of transitivity, reflexivity etc. I do not know of any operations on cardinalities, except for finite sets.

The abovementioned question can be solved if we know that $2^{(\mathbb N \times \mathbb N)}$ and $(2^\mathbb N)^{\mathbb N}$ have the same cardinality, as mentioned in one of the answers. I am unable to find a proof for the same.

Answer 1

HINT: Use a bijection from $\Bbb N\times\Bbb N$ to $\Bbb N$ to show that there is a bijection between ${}^{\Bbb N\times\Bbb N}2$ and ${}^{\Bbb N}2$. (I prefer the notation ${}^XY$ for the set of all functions from $X$ to $Y$.)

Added: Then construct the natural bijection between ${}^{\Bbb N}\left({}^{\Bbb N}2\right)$ and ${}^{\Bbb N\times\Bbb N}2$. You can define it in either direction, but I find it a little more natural to start with a function $f:\Bbb N\times\Bbb N\to 2$ and convert it to a sequence $\langle f_n:n\in\Bbb N\rangle$ of functions $f_n:\Bbb N\to 2$ by $f_n(m)=f(\langle ?,?\rangle)$?

Finally, use Cantor’s theorem.

Answer 2

The remaining portion of the question — a special case of showing that $\left(A^B\right)^C\equiv A^{B\times C}$ for all sets $A, B, C$ — can be handled by the classical notion of currying: a function $f:(B\times C)\mapsto A$ can be canonically thought of as a function of two arguments, $f(b,c)$ where the first argument is $b\in B$ and the second argument is $c\in C$, whereas a function $g: C\mapsto (B\mapsto A)$ is a function that takes a value $c\in C$ and returns a function $h:B\mapsto A$ which takes a value $b\in B$ and returns a value in $A$. But given such a function $g$ we can define $f:(B\times C)\mapsto A$ as $f(b,c)=\left(g(c)\right)(b)$, and contrariwise, given such an $f$ we can define a function $g$ as $g(c) = f(\_,c)$. Note that this proof also works in the case where $A,B,C$ are all finite sets, and gives a combinatorial proof of the classical identity $\left(a^b\right)^c=a^{bc}$.