Prove that $|A^{B×C}| = |(A^B)^C|$ I'm having trouble proving that $|A^{B×C}| = |(A^B)^C|$   , where  $M^N$ is the set of all the functions $f:N \to M$. 
My thoughts: to prove this, I need to find a bijection between $|A^{B×C}|$ and $|(A^B)^C|$, so I need a bijection between the set of all functions $g:B×C \to A$ and the set of all functions $h: C \to A^B$. 

The bijection actually should be between the set of all functions $g:B\times C \rightarrow A$ and the set of all functions $h:C\rightarrow A^B$ (otherwise you'd be proving that $|A^{B\times C}|=|A^C|$, which is false in general).

Added after the asker's edit:
In the wikipedia article of currying linked by Clive Newstead, there is a mention of the important and quite general phenomenon of "adjointness". Let's change the notation a little bit: Put $\hom(B,A)=A^B$. Then we are looking for bijections 
$\hom(B\times C,A)\cong\hom(C\times B,A)\cong \hom(C,\hom(B,A))$.
This is quite helpful when trying to remember how does the explicit bijection comes around: for a function $f:C\times B\rightarrow A$, fixing a coordinate $c_0\in C$ will yield a function $f(c_0,-):B\rightarrow A$, so $f$ can be seen as a function such that, when evaluated in a point $c_0$, yields a function from $B$ to $A$ (this of course is not a proof, rather a hint).
