Invertibility of $BA$ I'm having trouble with the following question which may seem simple but to me it's not

Let $A \in \mathbb{R}^{5 \times 7}, B\in \mathbb{R}^{7 \times 5}$.
Prove that $BA$ is not invertible.


I thought maybe use the relation to the solution sets of $Ax=0$ which is infinite, hence $A$ isn't invertible. But I don't know where to head from here?

By the way, I'm looking for a counterexample that shows $AB$ doesn't have to be non-invertible.
Thanks in advance to all who bother to reply!
 A: $BA$ is a $7 \times 7$ matrix, so it needs to be rank 7 to be invertible, But, since $\mbox{rank} (BA) \leq \min(\mbox{rank}(A),\mbox{rank}(B)) \leq 5$ (since $A,B$ have one dimension as $5$) it cannot be invertible. 
$AB$ on the other hand can be invertible (for example, generate two matrices with i.i.d. standard normal entries for $A,B$ and then multiply them together and you'll probably get an invertible matrix if you try it a few times). 
A: $A$ is a $5 \times 7$ matrix, so its null space must have dimension at least $2$.  If $Ax = 0$ then $BAx = 0$.  So the null space of $BA$ has dimension at least $2$, and $BA$ is not invertible.
A: The matrix $BA$ cannot be invertible: there exists a vector $x\in\mathbb{R}^7$ such that $Ax=0$ and $x\ne0$, because the homogeneous linear system $Ax=0$ has infinitely many solutions. Then $BAx=0$, which can't happen for an invertible matrix.
However, the matrix $AB$ can be invertible. Just to make a simple example,
$$
\begin{bmatrix} 1 & 0 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \end{bmatrix}
$$
is invertible. Now it's easy to extend this to the required size:
$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
