$A^k = 0$ means $A$ is a Nilpotent Matrix.
And all eigen values of a Nilpotent Matrix are $0$. So All, Eigen values of $(I+A)$ are $1$. So $\tt det(I+A)$ = Product of eigen values $= 1$
This can be seen from an example : Let $A = \begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$
$I+ A = \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$
So, $det(I+A) = 1$