If $B$ is nilpotent and $AB=BA$ then $\det(A+B)=\det(A)$ (Asking for other method)

Question

Let $K$ be some field and $A, B \in M_n(K)$. Prove that: If $B$ is nilpotent and $AB=BA$ then $\det(A+B)=\det(A)$.

I believe there is a nice solution here. However, it seems that this problem could be solved using Jordan Canonical Form of the matrices. I am wondering how. Could anyone give me a hint?