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!!!