prove that if a square matrix $A$ is invertible then $AA^T$ is invertible. prove that if a square matrix $A$ is invertible then $AA^T$ is invertible.
and also prove the opposite, that if $AA^T$ is invertible, then $A$ is invertible.
i wrote that $det(A) = det(A^T)$
and that $det(A) \neq 0$ when $A$ is invertible
and $det(A) = det(A^T) \neq 0$ 
and since product of invertible matrices are also invertible, then $AA^T$ must be invertible. 
but somehow I feel I will get told "I didn't do any work, I just said that the question was true" again like I did on my test. :| is there a perfect formal way I can prove this?
and also prove the opposite, that if $AA^T$ is invertible, then $A$ is invertible?

If $A$ is invertible then so is $A^T$ (with inverse the transpose of $A^{-1}$). The product $BC$ of two invertible matrices $B,C$ is invertible, with inverse $C^{-1}B^{-1}$. Now combine these facts. No determinants required.
For the converse, if a product of square matrices in invertible, then both factors must be invertible (since $C(BC)^{-1}$ is a right-inverse of$~B$ and $(BC)^{-1}B$ a left-inverse of $C$; for square matrices a one-sided inverse is automatically a two-sided inverse). Here maybe determinants could simplify the argument a bit: from $0\neq\det(BC)=\det(B)\det(C)$ it follows that $\det(B)\neq0$ and $\det(C)\neq0$.
