cardinality of $E^F$, $E, F = \emptyset$ My textbook (Naive Set Theory) asks the reader to show that $\left| E^F \right|$ = $\left| E \right| ^ \left| F \right|$ for all finite sets. In passing to induction, I noticed that this would imply that $0^0 = 1$, since there exists the trivial empty function (i.e., $\emptyset \times \emptyset = \emptyset$). The definition of exponentiation in $\omega$ says that $(\forall m \in \omega $ $(m^0 = 1))$, so this seems reasonable.  
Is this correct, or should this case be discarded? If it is correct, is it correct in $\omega$ only, or everywhere?
 A: As I mentioned when dealing with this extensively elsewhere, in set theory there is no reason to treat $0^0$ any different than any other set-theory exponentiation (set of all funtions from $\varnothing$ to $\varnothing$) or cardinal exponentiation (the cardinality of the set of all functions from $\varnothing$ to $\varnothing$). The general definition of exponentiation of sets, of cardinals (and ordinals, for that matter) applies to this special case and gives the value $1$. This value fits well with the corresponding uses within set theory.
There are other contexts, separate from set theory (e.g., the Calculus of limits on the real line) where one might wish to avoid giving a specific value to $0^0$; this is not one of them, and so you should feel absolutely no qualms in applying the general definition of exponentiation for sets/cardinals/natural numbers to that special case, in this context.
A: The empty product is equal to $1$.  To neglect to multiply by anything, is the same as multiplying by $1$.
Look at the well-known identity:
$$
e^z = \sum_{n=0}^\infty \frac{z^n}{n!}.
$$
If $z=0$, then this is
$$
e^0 = \frac{0^0}{0!}+0+0+0+\cdots.
$$
What should the first term be if $e^0=1$?
$0^0$ is an indeterminate form in the sense that if $f\to0$ and $g\to 0$ as $x\downarrow0$, then the limit of $f^g$ could be any non-negative number, depending on which functions of $x$ are $f$ and $g$.  But I think if they are analytic, then the limit will always be $1$.
