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)

Here is an alternative proof which avoids triangularization. The proof works over any field $k$.
Case 1:  $\det(A)=0$.
Then $\det(A+B)=0$, since otherwise $A+B$ is invertible, and then $A=(A+B)-B$ is invertible too (SE/119904), namely via the geometric series. This is a contradiction.
Case 2: $\det(A) \neq 0$.
Then $A$ is invertible. Then $C:=A^{-1} B$ is nilpotent with $\det(A+B)=\det(A) \det(1+C)$, so that it suffices to prove $\det(1+C)=1$. But this is clear since $0$ is the only eigenvalue of $C$ and therefore $1$ is the only eigenvalue of $1+C$. (In general, eigenvalues add when we add constants.)
