If $A$ is an $m\times n$ matrix, $B$ is an $n\times m$ matrix and $nThe question was given in the early chapters of Linear Algebra by Hoffman & Kunze, so I am trying to give a proof with only the tools given to me so far - which are mainly row reduction and knowledge of matrix multiplication, row reduced echelon forms, row equivalence and linear independence.
I attempted a proof as per the following:

Consider $A$ as a collection (not sure if this would be the ideal expression) of $1 \times n$ row vectors, and $B$ as a collection of $n \times 1$ column vectors. Then we have that:
$$
A=\begin{bmatrix} r_1 \\ \vdots \\ r_m \end{bmatrix},\ B=\begin{bmatrix} c_1 & \cdots & c_m \end{bmatrix}.
$$
  Thus it follows that:
  $$
AB =\begin{bmatrix} r_1\cdot c_1 & \cdots & r_1\cdot c_m \\ \vdots & ~ & \vdots \\ r_m\cdot c_1 & \cdots & r_m \cdot c_m  \end{bmatrix}
$$
  Clearly, by inspection, the rows are linearly dependent.
Since the rows of $AB$ are linearly dependent, it naturally follows that the reduced row echelon form of $AB$ contains zero rows. Hence, $AB$ is not invertible.

Would this be a mathematically sufficient proof?
 A: As has been mentioned in the comments, your approach does not make a for complete and proper proof. You may proceed as follows:
We have from here, for example, that $\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))$. This you could try to prove using the tools you already know.
In your problem we have that $\operatorname{rank}(A)\leq n$ and $\operatorname{rank}(B)\leq n$ implying that $\operatorname{rank}(AB) \leq n$, but $AB$ is $m$ by $m$ with $n<m$ hence $AB$ can not have full rank which in turn means that it is not invertible. 
A: Here's a proof that relies on matrix multiplication.
We can adjoin $m-n$ columns of zeros to $A$ and $m-n$ rows of zeros to $B$ to form $m\times m$ matrices $A', B'$. This won't affect the product $AB$, meaning $A'B'=AB$.
Then we'll have something like 
$$ \left[\begin{array}{c|c}  A & 0 \end{array}  \right] \left[\begin{array}{c} B\\ \hline 0 \end{array}\right]=A'B'$$
Since $A'$ has a column of zeros, $\det A'=0$, so $\det{A'B'}=0$ .
But since $AB=A'B'$, we have $\det{AB}=\det{A'B'}=0$, which means that $AB$ is not invertible.
