= Solution
Let $C=[A,B]$ be <trace> class and set $F(t)=e^{tA}e^Be^{-tA}e^{-B}$. The integral identity $e^BAe^{-B}-A=\int_0^1e^{sB}[B,A]e^{-sB}ds$ is trace-norm convergent. It shows that this difference is <trace> class. The analogous Duhamel identity then makes $F(t)-I$ <trace> class and differentiable in that <norm>. Hence $F(t)\in G(H)$.
Compute the right logarithmic <derivative> without taking separate traces of $A$:
$$
\dot F F^{-1}=e^{tA}(A-e^BAe^{-B})e^{-tA}.
$$
<Trace> invariance under bounded invertible similarity and the integral identity give $\operatorname{Tr}(\dot F F^{-1})=-\int_0^1\operatorname{Tr}[B,A]ds=\operatorname{Tr}[A,B]$. Cyclicity with a <trace-class> factor makes this equal to the <trace> in part (b). As $F(0)=I$, integrating the scalar differential equation gives
$$
\boxed{\det(e^Ae^Be^{-A}e^{-B})=\exp\operatorname{Tr}[A,B].}
$$
This <Fredholm determinant of an exponential commutator> can be nontrivial in infinite <dimension>: it is not legitimate to set the <trace> of a <trace-class> <commutator> to zero when $A,B$ themselves need not be <trace> class.
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