Prove that if matrix AA is invertible then A is invertible

Question

Suppose $AA$ is an invertible matrix, how would you prove that $A$ is invertible?

My Attempt: consider the equation $A\vec{x}=\vec{0}$ then $AA\vec{x}=A\vec{0}=\vec{0}$ since $AA$ is invertible then by the invertible matrix theorem since Nul(AA) ={$\vec{0}$} the unique solution for $\vec{x}$ is $\vec{0}$ so by the system Rank theorem and the fact that A is a squrae matrix rank(A)=n and thus A is invertible. Would this be a correct proof, or am I assuming to much by stating that $\vec{x}=\vec{0}$ is the only solution to $A\vec{x}=\vec{0}$?

Answer 1

Because $$\det{AA}\neq0$$ and $$\det{AA}=\left(\det{A}\right)^2\neq0,$$ which gives $\det{A}\neq0.$

Answer 2

By contradiction suppose $Ax=0$ for some $x\neq 0$ then $AAx=0$.

Answer 3

Let $B=A(AA)^{-1}$. What is $AB$?

Answer 4

Let $B=(AA)^{-1}$. Then $A(AB)=(AA)B=I$ and $(BA)A=B(AA)=I$. Thus $A$ has right and left inverse and hence $AB=BA$ is the inverse of $A$.