= Solution
Write $D=[A,B]$, so $[A,D]=[B,D]=0$. Repeated use of the product rule for the <commutator> gives $[A,B^n]=nDB^{n-1}$. Summing the exponential series yields
$$
\boxed{[A,e^B]=De^B.}
$$
These manipulations hold for <bounded operators>, or as identities on a common invariant domain where the series and derivatives are justified.
The same commutation relation implies $e^{\lambda A}Be^{-\lambda A}=B+\lambda D$: its derivative is $e^{\lambda A}[A,B]e^{-\lambda A}=D$ and its initial value is $B$. Put $G(\lambda)=e^{\lambda A}e^{\lambda B}$. Then
$$
G'=(A+B+\lambda D)G.
$$
For the specified $F(\lambda)=G(\lambda)e^{-\lambda(A+B)}$, the operator $G$ commutes with $A+B$. Indeed conjugation by $e^{\lambda B}$ sends $A+B$ to $A+B-\lambda D$, and subsequent conjugation by $e^{\lambda A}$ restores $A+B$. Therefore
$$
F'=(A+B+\lambda D)F-F(A+B)=\lambda DF,\qquad F(0)=I,
$$
so $F(\lambda)=e^{\lambda^2D/2}$. At $\lambda=1$ this proves the central-commutator case of the <Baker--Campbell--Hausdorff formula>:
$$
\boxed{e^Ae^B=e^{[A,B]/2}e^{A+B}=e^{A+B}e^{[A,B]/2}.}
$$
The last equality holds because $D$ commutes with $A+B$.
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