Prove that $$card(A^{B \times C})=card(A^{B^C})$$
where $A^B$ is a set of all functions from $B$ to $A$ and $A \times B$ is cartesian product of sets. Is the bijection supposed to be sending $f$ to $g$, where $g$ is sending $h$ to $t$, or something like that ? Can't work it out.
Note that with $A^{(B^C)}$, the claim is false - we want ${(A^B)}^C$ (which we recall to indeed equal $A^{BC}$ in the realm of natural numbers).
Given $f\colon B\times C\to A$, we want to find $g\colon C\to A^B$, i.e., $g$ should, for each $c\in C$, produce a map $g(c)\colon B\to A$. The choice $g(c)(b)=f(b,c)$ suggests itself. This correspondence invertible, i.e., given $g\in {(A^B)}^C$, we can define $f\colon B\times C\to A$ by letting $f(b,c)=g(c)(b)$.