prove if $AB$ is invertible, $A,B$ is also invertible. 
If $BA$ is invertible (where $A,B$ are matrix), then $A,B$ are invertible.

I want to prove this theorem by not using the fact that if $BA$ is invertible, then we know that $(BA)^{-1} = A^{-1}B^{-1}$, so there A, B are also invertible.
I want to use the theorem that $AX=0$ has only the trivial solution if and only if $A$ is invertible.
By using this fact, according to the condition that $BA$ is invertible, we can say $(BA)X=0$ has only the trivial solution. But I can't prove the next stage that $AX=0, BX=0$ has also only the trivial solution.
 A: It follows from the fact that
$$\text{rank}(BA) \leq \min\{\text{rank}(A), \text{rank}(B)\}$$
or
$$
\det(AB) = \det(A)\cdot\det(B)
$$
In fact in terms of linear mappings you have to prove that if $A \circ B$ is surjective then both $A$ and $B$ are surjective but it is obvious.
A: I assume that your matrices are square matrices.
You have that $BA$ is invertible, so $BAx = 0$ has only the trivial solution $x = 0$. Now say that $Ay = 0$ has a non trivial solution $y'$. Then $(BA)y' = B(Ay') = B0 = 0$. Hence $y'$ is also a non-trivial solution to $(BA)x = 0$. Hence (by contradiction) $A$ is invertible.
Now say that $By = 0$ has a non-trivial solution $y'$. Then let $y'' = A^{-1}y'$ and $y' = Ay''$. Then $y''$ is a non-trivial solution to $(BA)x = 0$: $BAy'' = BAA^{-1} y' = By' = 0$.
A: $$\begin{align}
BA\text{ is invertible} &\iff \det(BA)\neq 0\\
&\iff \det(B)\det(A)\neq 0 \\
&\iff \det(B)\neq 0\text{ and }\det(A)\neq 0\\
&\iff \text{$B$ is invertible and $A$ is invertible}
\end{align}
$$
A: you need the matrices $A, B\ $ to be square, otherwise as @GEdgar pointed out the result is not true.
suppose $A$ is not invertible, then there is an $x \neq 0$ such that $Ax=0.$ now, $BAx = B0 = 0$ and that $BA$ is invertible so $x = 0.$ contradicting our assumption. so
we have the square matrix $A$ invertible.
showing  $B$ is invertible. the column space of $BA$ is all of $R^n$ but column space of $BA$ is part of the column space of $B,$  therefore the column space of $B$ is all of $R^n.$ and that implies $B$ is invertible. 
