The existence of the set of all functions on A into B I need to prove that the set of all functions on A into B, denoted by $B^{A}$, exists.
I think if $|A| = m$ and $|B| = n$ then $|B^{A}|=n^{m}$ because for each element of $A$ there are $n$ independent ways of mapping it to $B$.
I don't know how to prove this.
the last time I posted this questions it got negative votes and closed.  
It is a question, asked in Hrbacek and Jechs book.  I need to answer it.  Please help.
 A: Since $A$ and $B$ exist, thus $A \times B$ exists, so $\mathcal{P}(A \times B)$ exists. Thus by the schema of separation (aka the schema of comprehension), we have that $$\{f \in \mathcal{P}(A \times B) : f \mbox{ is a function } A \rightarrow B\}$$
exists. But this is just $B^A$, by definition. So we're done.
Note that I am using the statement
$$f \mbox{ is a function } A \rightarrow B$$
as shorthand for the statement
$$\forall a \in A,\exists! b \in B : (a,b) \in f.$$
This usage, while convenient, is non-standard. In particular, in most other branches of mathematics we take $f : A \rightarrow B$ to mean that $f$ has source $A$ and target $B$.
A: Your argument doesn't explain why the set exists, and it doesn't suggest what to do if one of the sets is infinite.
Now that every function from $B$ to $A$ is a subset of $B\times A$. Show that we can define when a set is a function whose domain is $B$ (where $B$ is a parameter), then you can use separation to conclude the set of functions is a subset of $\mathcal P(B\times A)$.
