Proof by contrapositive with division 
Proof by Contraposition.  Include the contrapositive statement. If the product of two integers is not divisible by some integer $n$; then neither integer is divisible by $n$.

I know you negate  and reverse everything. So, the contrapositive statement would be: If both integers are divisible by $n$, then the product of two integers is divisible by some integer $n$.
Then I would do direct proof of this new statement.
I believe I would start off with:
Assume that there exists an integer $m$ and $p$ such that $m/n$ and $p/n$...
 A: You don't have it quite right. You need to show that if at least one of the two integers is divisible by $n$, then the product is.
So let $a$ and $b$ be integers, at least one of which is divisible by $n$. We show that $ab$ is divisible by $n$. 
Suppose that $a$ is divisible by $n$. Then $a=na'$ for some integer $a'$. Continue.
Added: Because by assumption $a$ is divisible by $n$, there exists an integer $a'$ such that $a=na'$. That's the meaning of divisibility. The number $a$ is divisible by $n$ if $a$ is $n$ times something, where the something is an integer. 
From the fact that $a=na'$, we conclude that
$$ab=(na')b=n(a'b).$$
The above equation says that $ab$ is equal to $n$ times the integer $a'b$. It follows that $n$ divides $ab$.
In a very similar way, we can show that if $n$ divides $b$, then $n$ divides $ab$. For if $n$ divides $b$, then $b=nb'$ for some integer $b'$. (Then proceed essentially like before.)
Remarks: $1.$ Because of the symmetry between $a$ and $b$, we can start as follows. Suppose that $n$ divides (at least) one of $a$ and $b$. Without loss of generality we may assume that $n$ divides $a$. Then continue as we did, without bothering to deal separately with the case $n$ divides $b$.
$2.$ The above is, unfortunately, abstract. Let's do a concrete example that may help. Let $a=75$ and $b=32$. We know that $n$ divides $75$, where $n=15$. Can we show that $n$ divides $(75)(32)$?  Well, we could do it the long way, by multiplying $75$ and $32$, and then seeing whether $15$ divides the product. But we can do it a shorter way. We have $75=(15)(5)$. So
$$(75)(32)=(15)(5)(32)=(15)](5)(32)].$$
Now we can see that $15$ divides $(75)(32)$. In fact we can see that the quotient is $(5)(32)=160$.
