TITLE: How to show $(a^b)^c=a^{bc}$ for arbitrary cardinal numbers?
QUESTION [23 upvotes]: One of the basic (and frequently used) properties of cardinal exponentiation is that $(a^b)^c=a^{bc}$.
What is the proof of this fact?
As Arturo pointed out in his comment, in computer science this is called currying.

REPLY [19 votes]: Martin gave a complete and detailed answer, I will give a short proof that might be clearer.
Recall that $A^B = \{f\mid f\colon B\to A\}$, i.e. all the functions from $B$ into $A$.
Since we defined cardinality as equivalence classes, it suffices to show that $(A^B)^C$ is equinumerous with $A^{B\times C}$ for arbitrary sets.
First we need to understand these two sets. $(A^B)^C$ is a set of functions, whose domain is $C$ and their value is a function from $B$ into $A$. That is a typical function would take an element of $C$ and return a function from $B$ into $A$.
$A^{B\times C}$ is the set of all functions whose domain is $B\times C$, into $A$. These functions take ordered pairs (or two variables) from $B\times C$ and return an element of $A$.
The intuition of what we are going to do is similar to $f_n = \cos (nx)$ and $f(x,n) = \cos (nx)$. The former is a function from $\mathbb N$ into $\mathbb R^\mathbb R$ (namely $n\mapsto \cos(nx)$), while the latter is a function from $\mathbb R\times\mathbb N\to\mathbb R$.
The way we do it is by defining $\varphi\colon(A^B)^C\to A^{B\times C}$ defined as in a similar way. $\varphi(f)$ will be the function that when fixing $c$ we have $f(c)$.
Rigorously, this is defined as: $$\varphi(f)(b,c) = f(c)(b)$$
To check that this is one-to-one, if $g,f\in (A^B)^C$ are two different functions, take $c$ such that $f(c)\neq g(c)$ (remember that both of these values are functions from $A^B$). That means that for some $b\in B$ we have $f(c)(b)\neq g(c)(b)$.
Therefore at the pair $\langle b,c\rangle$ the function $\varphi(f)$ will have a different value than $\varphi(g)$.
To show that $\varphi$ is a surjection, we take a function $g\colon B\times C\to A$, and simply set $f\colon C\to A^B$ to be $f(c)(b) = g(b,c)$. We need, of course, to show that $f$ is well defined, and then we can easily show that $\varphi(f)=g$.