An identity for nested commutators Let $A,B$ be Hermitian operators on an arbitrary Hilbert space. Define nested commutators of $B$ with respect to $A$ as $\text{Ad}_{A}(B) = \left[A,B\right]$, $\text{Ad}_{A}^{2}(B) = \left[A,\left[A,B\right]\right]$, and so on, i.e., $\text{Ad}_{A}^{k}(B)$ contains $k$ such nested commutators. Someone recently mentioned to me that the following identity holds for arbitrary integers $k$, $q$:
$$\text{Tr}\left[\text{Ad}_{A}^{k}(B)\text{Ad}_{A}^{q}(B)\right] = -\text{Tr}\left[\text{Ad}_{A}^{k+1}(B)\text{Ad}_{A}^{q-1}(B)\right]$$
I think this identity is really neat, but I don't have a clean proof of it and I haven't seen it elsewhere. Does anyone have a neat way to prove this? This seems like the kind of identity which could make life a lot easier sometimes.
 A: It's quite a straightforward proof if we use the linearity $\operatorname{Tr}(A + B) = \operatorname{Tr}(A) + \operatorname{Tr}(B)$ and cyclic invariance $\operatorname{Tr}(ABC) = \operatorname{Tr}(CAB)$ of the trace. I'll write $\mathcal{A} := \operatorname{Ad}_A(B)$ for brevity.
$$
\operatorname{Tr}\big(\mathcal{A}^k\mathcal{A}^q\big) = \operatorname{Tr}\big(\mathcal{A}^k [A, \mathcal{A}^{q-1}]\big) \\
= \operatorname{Tr}\big(\mathcal{A}^k(A \, \mathcal{A}^{q-1} - \mathcal{A}^{q-1}\,A)\big)\\
= \operatorname{Tr}\big(\mathcal{A}^k \, A \, \mathcal{A}^{q-1}\big) - \operatorname{Tr}\big(\mathcal{A}^k \, \mathcal{A}^{q-1} \, A\big)\\
= \operatorname{Tr}\big(\mathcal{A}^k \, A \, \mathcal{A}^{q-1}\big) - \operatorname{Tr}\big(A \, \mathcal{A}^k \, \mathcal{A}^{q-1}\big)\\
= \operatorname{Tr}\big(\mathcal{A}^k \, A \, \mathcal{A}^{q-1} - A \, \mathcal{A}^k \, \mathcal{A}^{q-1}\big)\\
= \operatorname{Tr}\big((\mathcal{A}^k \, A - A \, \mathcal{A}^k)\mathcal{A}^{q-1}\big)\\
= \operatorname{Tr}\big(-[A, \mathcal{A}^k]\mathcal{A}^{q-1}\big)\\
= -\operatorname{Tr}\big(\mathcal{A}^{k+1}\mathcal{A}^{q-1}\big)\quad\square\\
$$
