The empty function and constants On Wikipedia and also after searching this forum there is stated that for each $A$ there exists a unique function $f : \emptyset \to A$ called the empty function for $A$.
In the theory of Algebraic Data Types (ADT's) and in functional programming languages (like Haskell) where one paradigma is "everything is a function" it is common to interpret constants as $0$-ary functions. A nullary function is just a function of the form $f : \emptyset \to A$, and so for every $A$ there must be such a function under this interpretation (so $|A|$ such functions) and this contradicts the uniques of one empty function for every $A$.
Also on Wikipedia I read.

The n-ary cartesian power of a set X is isomorphic to the space of functions from an n-element set to X. As a special case, the 0-ary cartesian power of X may be taken to be a singleton set, corresponding to the empty function with codomain X.

So $X^0 = \{ x \}$, but otherwise if I interpret $Y^X$ as the set of functions from $X$ to $Y$ and interpret $X^n$ as $X^{\{1,\ldots,n\}}$ then $X^0 = X^{\emptyset} = \{ f : \emptyset \to X \}$? Could someone please clarify, I am confused...
EDIT: Okay guess I got it now, but still confused. For every $A$ there is a  unique $f : \emptyset \to A$, but that is not equal to $f : A^0 \to A$, with $A^0 = \{ f : \emptyset \to A \}$ there are exactly $|A|$ functions $f : A^0 \to A$. But then I guess the following one statement about the nullary product on Wikipedia is wrong, that if the index set $I$ is empty, that
$$
 \prod_{\emptyset} = \{ f_{\emptyset} : \emptyset \to \emptyset \}.
$$
because there the set over which the product is built is lost? 
 A: An $n$-any function on a set $X$ is a function $X^n\to X$, where $X$ is the $n$-fold cartesian product of $X$ with itself. Thus, a $0$-ary function is a function $X^0\to X$, not $\emptyset \to X$. So the question is to figure out what $X^0$ is. One way to reason about what it should be is to note that $X^n\times X^m$ is essentially the same as $X^{n+m}$ (as most people will agree is true for all $m,n>0$. To make this true also for $n=0$, we need, e.g., that $X^0\times X^m$ is essentially the same as $X^m$. Which set has $Y$ has the property that $Y\times X^m$ is essentially just $X^m$ (for $X\ne \emptyset$)? Well, a moment's thought should reveal that the answer is that $Y$ can be any singleton set. 
So, to preserve some basic realizations about the cartesian product of sets, it makes sense to define $X^0$ (for nonempty $X$) to be a singleton set (whichever one you want). 
Another way to argue is categorically. The cartesian product of sets is a special case of the notion of categorical product, and $X^0$ corresponds to an empty product. The universal property for the empty product is just a terminal object in the category. The terminal objects in the category of sets are precisely the singletons.
I just saw your edit: Notice that the conventions agree: $X^0=X^\emptyset =\{\emptyset \to X\}$ is a singleton set.   
A: A function $f$ from set $X$ to set $A$ is a triple $\left(X,G,A\right)$
where $G\subset X\times A$ and such that $\forall x\in X\exists!a\in A\;\left(x,a\right)\in G$.
If here $X=\emptyset$ then automatically $G=\emptyset$ and the mentioned
condition is satisfied vacuously. 
So $\left(\emptyset,\emptyset,A\right)$is
a function from $\emptyset$ to $A$ and is automatically unique.
Your definition $A^{0}=A^{\emptyset}=\left\{ f\mid f:\emptyset\rightarrow A\right\} $
is correct. It contains for every $A$ (not every element of $A$) exactly one element. 
Elements
of $A$ can be identified with functions $f:*\rightarrow A$ where
$*$ denotes a fixed singleton (not the empty set). Taking for instance
$*=\left\{ 0\right\} $ we have the function $\left(\left\{ 0\right\} ,\left\{ \left(0,a\right)\right\} ,A\right)$
corresponding with element $a\in A$. 
EDIT
We can define $\sqcap_{i\in I}X_{i}$ as the set of functions $f:I\rightarrow\cup_{i\in I}X_{i}$
with $\forall i\in I\; f\left(i\right)\in X_{i}$, but doing so it
must be kept in mind that here we are not defining the product, but a
product of the family of $\left(X_{i}\right)_{i\in I}$ and this with
projections $p_{i}$ defined by $f\mapsto f\left(i\right)$. If we
loose that out of sight than ambiguity can arise. For instance if
$\forall i\in I\; X_{i}=X$ then the product can be abreviated as
$X^{I}=\mathbf{Set}\left(I,X\right)$ . If $I\neq\emptyset$ then
this set equals the set $\sqcap_{i\in I}X_{i}$ defined as above.
However in special case $I=\emptyset\wedge X\neq\emptyset$ it does
not, because $\cup_{i\in I}X_{i}=\emptyset\neq X$. As magma states
well in his comment: products are defined up to isomorphism. In category
$\mathbf{Set}$ any singleton can serve as product of an empty family.
The singletons are in fact the terminal objects there.
A: A function from a set $A$ to a set $B$ is a subset $S$ of $A\times B$ such that for every $a \in A$, there is a unique $(a,b) \in S$. According to this definition, the empty subset $\phi$ of $\phi\times B$ is a function, and $\phi$ is the unique function from $\phi$ to $B$. Therefore, the number of functions from $A$ to $B$ (for finite sets $A,B$) is $|B|^{|A|}$, which is also consistent with the fact that the number of functions from $\phi$ to $A$ for any finite $A$ is $|A|^0 = 1$.
