If $A \ne \varnothing$, then $\varnothing^A=\varnothing$ 
Prove that if $A\ne\varnothing$, then $\varnothing^A=\varnothing$. Here, $\varnothing^A=\{f\in \mathcal P(A\times \varnothing): f:A \to \varnothing\}$ is the set of functions from $A$ to $\varnothing$.
What if $A = \varnothing$?

Proof: Assume $f \in \varnothing^A$
Then $f \subseteq A \times \varnothing = \varnothing$
Thus $f = \varnothing$
Now we are trying to show that $\varnothing \in \varnothing^A$
$\varnothing \subseteq A \times \varnothing$. Thus $\varnothing \in \mathcal P(A \times \varnothing)$.
and $\operatorname{dom}(\varnothing)=\varnothing=A\times \varnothing$
And $\varnothing$ is a function, since if not then there exist $x,y,z$ such that $(x,y), (x,z)\in \varnothing$ and $y\ne z$. But this contradicts that $\varnothing$ has no elements. Thus $\varnothing$ is a function.

$\varnothing^\varnothing=\{\varnothing\}$
Am I doing it right?
 A: All of your steps are individually correct. However, they do not provide a solution to the problem yet.
Yes, $f = \varnothing \subseteq A \times \varnothing$ is a function.
But, it is not a function $f: A \to \varnothing$. For, to be a function with domain $A$, a relation (in addition to the functionality condition you mention in your question) has to be left-total:
$$\forall a \in A: \exists b: (a,b) \in f$$
But, since $A$ is nonempty, this statement is actually false for all $a \in A$; therefore, $f$ is not a function $A \to \varnothing$.
We conclude that $\varnothing^A = \varnothing$.

Indeed, if $A$ is empty, then any statement of the form $\forall a \in A: \phi(a)$ is vacuously true, since there are no elements in $A$ that could possibly make the statement false.
A: We always regard $B^A$ to mean the set of functions from $A$ into $B$.  Thus, if $A$ is not empty but $B$ is, then there are no such functions because let $a \in A$ and let $f:A\rightarrow B$.  Then $f(a) \in B$.  Curiously enough, this is not true when $A$ is empty, which means that this formalism is sort of pointing towards one of the conventions for what $0^0$ should be when it's defined on the real numbers.
