How to prove the equality of determinant? Let A be a square matrix, and B a nilpotent. (size of A and B is same)
Assume AB=BA
Show that, det(A+B)=det(A)

Hint: Since they commute, they are simultaneously upper triangularizable.  For some unitary $U$, we can write
$$
UAU^* = \pmatrix{\lambda_1&&&&*\\&\lambda_2\\&&\ddots\\&&&&\lambda_n}: = T_A\\
UBU^* = \pmatrix{0&&&&*\\&0\\&&\ddots\\&&&&0} := T_B
$$
Note that $\det(A + B) = \det(T_A + T_B)$
