1. Prove that [A,B^n] = nB^{n-1}[A,B]
Given that: [[A,B],B] = 0
My Atempt to resolution
We can write that:
[[A,B],B] = [A,B]B-B[A,B] = 0
So we get that: [A,B]B = B[A,B]
After some working several expansions, and considering that [X,YZ] = Y[X,Z] + [X,Y]Z
I arrived...