Is it always true for $\det(I+AA^*)\neq 0$ ?
Here, $A^*$ is the adjoint of $A$.
I have found that $\det(I+BC)=\det(I+CB)$, but it seems no use here.
Suppose $\det(I + A A^*) = 0$.
That means the quadratic matrix $B = I + A A^*$ is singular. Thus there exists a nonzero vector $v$ such that $B v = 0$, and therefore $v^* B v = 0$.
But $0 = v^* B v = v^* v + (A^* v)^* (A^* v) = \|v\|^2 + \|A^* v\|^2$ implies $v = 0$.
This is a contradiction.
Saying that $\det(AA^*+I)=0$ is the same as saying that $-1$ is an eigenvalue of $AA^*$; however, if $AA^*=\lambda v$ for some $v\ne0$, we have $$ (A^*v)^*(A^*v)=v^*AA^*v=\lambda(v^*v) $$ and, for any vector $x$, $x^*x\ge0$. Thus $\lambda\ge0$.
Another way (sketch)
If $P>0$ and $Q\ge 0 $ then $P +Q>0$ (here $>0$ means "strictly positive definite" and $\ge 0$ means "(semi) positive definite".
$P>0 \implies |P| \ne 0$
$A A^* \ge 0$
For any $\epsilon>0$, $\epsilon I>0$. Hence, $\epsilon I + A A^* > 0$
Therefore, $|\epsilon I + A A^*| \ne 0$