Determinant of a nilpotent matrix
Since $A$ is nilpotent, we have
$A^m = 0 \tag{1}$
for some positive interger $m$. This implies every eigenvalue of $A$ vanishes, since the equation
$Av = \lambda v \tag{2}$
for non-zero $v$ (recall eigenvectors are required to be non-zero) implies
$0 = A^mv = \lambda^m v, \tag{3}$
whence
$\lambda^m = 0, \tag{4}$
since $v \ne 0$. (4) forces
$\lambda = 0 \tag{5}$
Now use the fact that for any scalars $\lambda$ and $a$, $\lambda$ is an eigenvalue of $A$ if and only if $\lambda + a$ is an eigenvalue of $A + aI$; indeed we have, from (2),
$(A + aI)v = Av + av = (\lambda + a)v. \tag{6}$
(6) allows us to conclude that every eigenvalue of $A + I$ is $1$; hence $\det (A+I)$, being the product of its eigenvalues, satisfies
$\det(A+I) = 1. \tag{7}$
QED.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!