Prove $[A,B^n] = nB^{n-1}[A,B]$ I am trying to show that $[A,B^n] = nB^{n-1}[A,B]$ where A and B are two Hermitian operators that commute with their commutator. However, I am running into a little problem and would like a hint of how to proceed.
If A and B commute then $[A,B] = ABA^{-1}B^{-1} = e$ where e is the identity element of the group.
$$\therefore AB=BA$$
$$n=1; [A,B^1] = (1)B^0[A,B] = e$$
This statement is certainly true. however moving on to $n = 2$ I find...
$$[A,B^2] = AB^2A^{-1}B^{-2} = ABBA^{-1}B^{-1}B^{-1} = BBAA^{-1}B^{-1}B^{-1}$$
Where in the last step I have used the fact that A and B commute to rearange the terms. However, it is plain to see that this last term simply reduces to the identity as well and for the n = 2 case we have:
$$[A,B^2] = e \ne 2B[A,B] = 2Be = 2B$$
Clearly I have assumed something I shouldn't have. The fact that there is a multiplicative factor of n implies I should be adding things, but I thought if I kept it as general as possible, the answer should just fall out naturally. I don't want an answer please, only guidance.
 A: Here's another way to prove this relationship, by induction:


*

*Verify that the statement holds for $n = 1$


$$
[A,B^1]=1 \cdot B^{1-1} [A,B] = [A,B]
$$ 


*Show that, if the formula holds for $n=k ~ \big(I\big)$, then it also holds for $n=k+1$, using the identity $[X, YZ] = [X,Y]Z + Y[X,Z] ~\big(II\big)$ and the fact that $B$ commutes with $[A,B] ~\big(III\big)$ 


$$
\begin{align}
[A,B^{(k+1)}] &= [A,B^k B] \\
& = [A,B^k]B + B^k[A, B] ~\big(II\big)\\
& = k B^{k-1}[A,B]B + B^k[A, B] ~\big(I\big) \\ 
& = k B^{k-1}B[A,B] + B^k[A, B] ~\big(III\big) \\
& = k B^k[A,B] + B^k[A, B] \\
& = (k + 1) B^k[A,B] \\
& = (k + 1) B^{(k+1)-1}[A,B] \\
\end{align}
$$


*Since both the basis and the inductive step have been performed, by mathematical induction, the statement holds for all natural numbers $n$. $\Box$

A: Maybe you could show this relation in an other way:
$[A,B^n]=[A,B^{n-1}B]=B^{n-1}[A,B]+[A,B^{n-1}]B=...$
Then you take $[A,B^{n-1}]B$ and repeat the process:
$[A,B^{n-1}]B=[A,B^{n-2}B]B=B^{n-2}[A,B]B+[A,B^{n-2}]BB$
So $B$ commute with $[A,B]$
$[A,B^{n-1}]B=B^{n-1}[A,B]+[A,B^{n-2}]B^2$
Repeat for $n$ steps
$[A,B^n]=B^{n-1}[A,B]+B^{n-1}[A,B]+...+B^{n-1}[A,B]=nB^{n-1}[A,B]$
A: It seems that the question (v1) is caused by the fact that there are two different notions of the commutator: 


*

*One for group theory: 
$$\tag{1} [A,B] ~:=~ ABA^{-1}B^{-1}$$ 
(or sometimes $[A,B] := A^{-1}B^{-1}AB$, depending on convention), which is relatively seldom used in physics.

*One for rings/associative algebras: 
$$\tag{2} [A,B]:=AB-BA,$$ 
which is the definition usually used in physics. (This latter definition (2) generalizes to a supercommutator in superalgebras.)
The identity 
$$\tag{*}  [A,B^n] ~=~ nB^{n-1}[A,B]$$ 
holds in the latter sense (2), if $[[A,B],B]=0$. (It is not necessary to demand $[A,[A,B]]=0$.) More generally, for a sufficiently well-behaved function $f$, we have
$$\tag{**} [A,f(B)] ~=~ f^{\prime}(B)[A,B], $$
if $[[A,B],B]=0$. 
The group commutator (1) is dimensionless, which (among other things) makes the identity (*) unnatural to demand for group commutators.
