Prove that if $A$ is invertible matrix then $AA^T$ and $A^T A$ are also invertible. I want to know how to prove this question:

Prove that if $A$ is invertible matrix then $AA^T$ and $A^T A$ are also invertible. 

My attempt: 
Since $A$ is invertible we have that $AA^{-1} = I$ and if we denote $B = A^{-1}$, we have that $AB=I$ so if we take the transpose of both sides then we have $(AB)^T = I^T = I$, but this is where I get my problem since the transpose, inverse of $B$ isn't equal to $A$. 
 A: Using your notation, you've shown that $(AB)^T = I$, so far so good; but also $(AB)^T = B^TA^T = (A^{-1})^TA^T$, and this tells you that $A^T$ is invertible and $(A^T)^{-1}=(A^{-1})^T$. Can you see where to go with this?
A: Recall a matrix is invertible if and only if its determinant is non-zero. 
Now what can you say about
A) $\text{det}(A^T)$
B) $\text{det}(AB)$?
Work these out, and it becomes clear
A: Let B be a matrix, so that $AB = I$ and $BA = I$. Then $A^T B^T = (BA)^T = I^T = I$ and $B^T A^T = (AB)^T = I^T = I$. So the inverse for $A^T$ is $B^T$. Now we get :
$(A A^T)(B^T B) = A(A^T B^T) B = A(I)B = AB = I$ and $(A^T A)(B B^T) = I$. So the inverse for $AA^T$ is $B^TB$ and for $A^T A$ we have $B B^T$.
A: An alternative approach:
To show that $A^TA$ is invertible, think about the linear system $A^TAx=0$. It suffices to show that $x=0$ is the unique solution. But $A^TAx=0$ implies that $\langle Ax,Ax\rangle= x^TA^TAx=0$. Thus $Ax=0$. Now use the fact that $A$ is invertible.
For $AA^T$, note that  $AA^T=(A^T)^T(A^T)$ 
A: last line $(A^TA)^T=A^TA$. 
$AA^T$ is equivalent to $(A^T)^T(A^T)$, so the rest follows when $A^T$ is treated as A.
A: You need to find an $\;X\;$ such that $\;A A^T X = I\;$, given that $\;A\;$ is invertible.
Let's calculate what this means, by repeatedly left-multiplying by a matrix that cancels out:
\begin{align}
(*) \;\;\; \phantom{\equiv} & A A^T X = I \\
\Rightarrow & \;\;\;\;\;\text{"left-multiply by $\;A^{-1}\;$; use $\;A^{-1} A = I\;$"} \\
& A^T X = A^{-1} \\
\Rightarrow & \;\;\;\;\;\text{"left-multiply by $\;(A^{-1})^T\;$"} \\
& (A^{-1})^T A^T X = (A^{-1})^T A^{-1} \\
\equiv & \;\;\;\;\;\text{"simplify using $\;P^T Q^T = (Q P)^T\;$ and $\;A A^{-1} = I\;$"} \\
(**) \;\;\; \phantom{\equiv} & X = (A^{-1})^T A^{-1} \\
\Rightarrow & \;\;\;\;\;\text{"left-multiply by $\;A A^T\;$ -- working our way back to $(*)$"} \\
& A A^T X = A A^T (A^{-1})^T A^{-1} \\
\equiv & \;\;\;\;\;\text{"simplify using $\;P^T Q^T = (Q P)^T\;$, $\;A^{-1} A = I\;$, and $\;A A^{-1} = I\;$"} \\
& A A^T X = I \\
\end{align}
This 'ping pong' argument shows that $(*)$ and $(**)$ are equivalent, so $\;(A^{-1})^T A^{-1}\;$ is the inverse of $\;A A^T\;$.
The other half of the question can be solved in the same way.
