Need to prove that there are $m^{n}$ mappings between two sets. Is my proof formal enough, and if not, what can I improve on? Here is the proposition:
$\textbf{Proposition.}$ There are $m^{n}$ mappings between a set $A$ and a set $B$, where $|A|=n$ and $|B|=m$.
$Proof.$ Let $A$ and $B$ be sets with $|A|=n$ and $|B|=m$. Let $a_{i}\in A$. Take this arbitrary element in A and observe that it can be mapped to $m$ different elements, as in, there are $m$ choices for this element. By the rule of product, the amount of mappings from $A$ to $B$ with any two elements in $A$ will be $m\cdot m=m^{2}$, as there are $m$ choices for every element. Continually applying the rule of product yields $m\cdot m \dotsm m\cdot m=m^{n}$. Thus, there are $m^{n}$ mappings between two sets with cardinality $n$ and $m$.
How can I improve this proof and make it more verbose and formal?
 A: That proof is  fine. As an instructor, it's just what I would hope to see.  Making it "more verbose and formal" would only make it harder to read and understand.
A: We need a few preliminary results:

*

*For any finite sets $A$ and $B$ and binary partition $\{A', A''\}$ of $A$, $\left|B^A\right| = \left|B^{A'}\right| \times \left|B^{A''}\right|$.

*For any singleton $A$ and finite set $B$, $\left|B^A\right| = |B|$.

Now, let $K$ be the set of natural numbers $k$
such that, for every set $A$ of size $k$, $\left|B^A\right| = m^k$.
Then, by induction, $K = \mathbf{N}$, since:

*

*For any set $A$ of size $0$ (that is, for the empty set), $B^A$ has just one element, the empty mapping, so
\begin{align*}
\left|B^A\right| &= 1\\
&= m^0;
\end{align*}
so $0 \in K$.


*Let $k$ be a member of $K$. Let $A$ be a set of size $k^+$.
Then, for some objects $A'$ and $a$,
$A$ is a set of size $k$ and $\{A', \{a\}\}$ is a partition of $A$.
Then
\begin{align*}
\left|B^A\right|
&= \left|B^{A'}\right| \times \left|B^{\{a\}}\right| && \text{by prelim. result 1}\\
&= \left|B^{A'}\right| \times |B| && \text{by prelim. result 2}\\
&= m^k \times m && \text{by inductive hyp.}\\
&= m^{k^+}.
\end{align*}
Since that holds for all sets $A$ of size $k^+$, $k^+ \in K$.
That is, for every element $k$ of $K$, $k^+ \in K$.
Now, since $n \in \mathbf{N}$ and $\mathbf{N} = K$, $n \in K$,
so, by construction of $K$, $\left|B^A\right| = m^n$.
