The main thing is presumably that $AA^T$ is symmetric. Indeed $(AA^T)^T=(A^T)^TA^T=AA^T$. For symmetric matrices one has the Spectral Theorem which says that we have a basis of eigenvectors and every eigenvalue is real.
Moreover if $A$ is invertible, then $AA^T$ is also positive definite, since $$x^TAA^Tx=(A^Tx)^T(A^Tx)> 0$$
Then we have: A matrix is positive definite if and only if it's the Gram matrix of a linear independent set of vectors.
Last but not least if one is interested in how much the linear map represented by $A$ changes the norm of a vector one can compute
$$\sqrt{\left<Ax,Ax\right>}=\sqrt{\left<A^TAx,x\right>}$$
which simplifies for eigenvectors $x$ to the eigenvalue $\lambda$ to
$$\sqrt{\left<Ax,Ax\right>}=\sqrt \lambda\sqrt{\left<x,x\right>},$$
The determinant is just the product of these eigenvalues.
There may be a difference between $A^{\star}A$ and $AA^{\star}$ when it comes to eigenvalue $0$, but that won't affect the spectral radius. For $\lambda \ne 0$, assume that $A^{\star}Ax=\lambda x$ for some $x \ne 0$. Now apply $A$ to both sides:
$$
AA^{\star}[Ax] = \lambda [Ax].
$$
So $\lambda$ is an eigenvalue because $Ax \ne 0$; indeed, if $Ax$ were $0$, then $A^{\star}Ax=\lambda x$ would be $0$ and, hence, $x=0$ would hold because $\lambda \ne 0$, which is a contradiction. This works the other way around, too. So the non-zero eigenvalues of $AA^{\star}$ and $A^{\star}A$ are the same.
Best Answer
No they're definitely not. Here's an easy counterexample:
$$ A=\left( \begin{matrix}1 & 1 \\ 0 & 0 \end{matrix} \right) $$ This matrix has eigenvalues $1$ and $0$ but $$ A^T A = \left( \begin{matrix}1 & 1 \\ 1 & 1 \end{matrix} \right) $$ which has eigenvalues $2$ and $0$.
It is true if $A$ is a normal matrix (i.e. $A^T A = AA^T$) though.