$[A^n,B]$ in terms of nested commutators

Question

For a project I am working on, it would be helpful to have an expression for $[A^n,B]$ for $[A,B] = AB-BA \neq 0$, in terms of nested commutators $[A,[A,B]]$ and so forth, with any $A$'s all moved to the left. This is because the nested commutators are trivial for me to calculate in my case. So far I have been able to derive:

$$[A^2,B] = 2A[A,B] - [A,[A,B]]$$

and that,

$$[A^3,B] = 3A^2[A,B] - 3A[A,[A,B]] + [A,[A,[A,B]]].$$

It is in theory easy to compute $[A^n, B]$ for a specific $n$, you write down all nestings up to $n$ nestings, and for each term you include the necessary factors of $A$ to the left, so that you have $n$ factors of $A$ in every term. However this is tedious, and does not provide a general formula. Is there a general expression for $[A^n,B]$ in the form I have derived above?

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Update

As a comment pointed out, the first two examples hinted at binomial coefficients, and I am conjecturing that,

$$[A^n,B] = \sum_{k=1}^n (-1)^{k-1} \binom{n}{k} A^{n-k} \underbrace{[A,[A,\dots]]}_{k\; \text{times}}.$$

Here $k=1$ nestings means the commutator $[A,B]$. I have checked this for $n=4$ by hand, and it continues to hold as well as for the other examples of course. I do not know how to prove it though.

Answer 1

Consider the two operators $L_A(B)=AB$ and $R_A(B)=BA$. These two operators commute with each other, so by the binomial theorem, $$R_A^n=(L_A-(L_A-R_A))^n=\sum_{k=0}^n (-1)^k \binom{n}{k} L_A^{n-k}(L_A-R_A)^k,$$ and thus $$L_A^n-R_A^n=\sum_{k=1}^n(-1)^{k-1}\binom{n}{k} L_A^{n-k}(L_A-R_A)^k.$$ This is exactly your conjectured formula since $L_A-R_A$ is the commutator operator $[A,-]$ and $L_A^n-R_A^n$ is $[A^n,-]$.

Answer 2

Let $\text{ad}_A(X)=[A,X]$. We can easily find a recursive formula for $[A^n,B]$: $$[A^n,B]=A^nB-BA^n$$ $$=A^{n-1}BA-BA^n+A^{n-1}[A,B]$$ $$=[A^{n-1},B]A+A^{n-1}[A,B]$$ $$=A[A^{n-1},B]-[A,[A^{n-1},B]]+A^{n-1}[A,B]$$ $$=A[A^{n-1},B]-\text{ad}_A[A^{n-1},B]+A^{n-1}[A,B]$$ It is straightforward to check that your conjectured closed formula $$[A^n,B] = \sum_{k=1}^n (-1)^{k-1} \binom{n}{k} A^{n-k} \text{ad}_A^k(B)$$ indeed satisfies this recursion, which proves it.