If $A$ and $B$ are $n\times n$ matrices, and $AB$ is invertible then $A$ and $B$ are invertible.
I started out by writing that since $AB$ is invertible, then for the equation $ABx=b$ has a unique solution for any $b$, $x=(AB)^{-1}b$. Then I don't know how to break up the composition of AB to prove they are each individually similar.
Remember that a square $A$ matrix is invertible if and only if $Ax=0$ iff $x=0$
Suppose that $AB$ is invertible and that $B$ isn't invertible, so then there is a $x\neq0$ such that $Bx=0$. Then $ABx=A(Bx)=A0=0$, with $x\neq0$, which is a contradiction since $AB$ is invertible. Hence $B$ is invertible.
Now suppose $A$ is not invertible, then there is $x\neq0$ such that $Ax=0$. Since $B$ is invertible, take $y=B^{-1}x$, where $y\neq0$ since $x\neq0$. Then $ABy=A(By)=Ax=0$, which is a contradiction. Hence $A$ is invertible.
$A,B$ are invertible if and only if $\det(A)\ne0$ and $\det(B)\ne0$, if and only if $\det(A)\det(B)=\det(AB)\ne0$, if and only if $AB$ is invertible.
Another simple way: For any two matrices $A,B$:
(*) $\text{rank}(AB) \le \min{\left( \text{rank}(A), \text{rank}(B)\right)}$
If $AB$ is invertible, then clearly $\text{rank}(AB)=n$. Combining with (*) gives: $n \le \min{\left( \text{rank}(A), \text{rank}(B)\right)}\le n$ and therefore $\text{rank}(A)=\text{rank}(B)=n$. Hence, both $A$ and $B$ are invertible.
Another way to show this is to keep in mind that :
$ \forall A \in M_n(K) $ ; A is invertible iff $ \exists B \in M_n(K) $ s.t $ AB = I_n $.
because for any matrix M : M is right invertible iff M is left invertible iff M is invertible. So, AB invertible implies that there exists a matrix C such that $ABC = I_n $
In other words:
$ A(BC) = I_n $ (by associativity of matrix product)
Thus A is invertible, We use invertibility at left to prove that B is invertible.
