Here I will give some algebra method for the proof of Glauber's formula:
Assume $F(t)=e^{At} e^{Bt}$ :
$$
\dfrac{d}{d t} F(t) = A e^{A t} e^{B t} + e^{A t} B e^{B t} = (A+e^{A t}Be^{-At} ) F(t) \tag{1}
$$
Recall that the Hadamard's lemma (Proved in appendix):
$$
\boxed{e^{A t} B e^{-A t} = B+ t[A,B]+\dfrac{t^2}{2!}[A,[A,B]]+\dfrac{t^3}{3!}[A,[A,[A,B]]]+\cdots} \tag{2}
$$
then the equation $(2)$ can be simplified as ($[A,B]=Constant$) :
$$
\dfrac{d}{d t} F(t) = (A+B+t[A,B])F(t) \tag{3}
$$
Assume $G(t) = e^{At+Bt+f(t)H(A,B)}$ :
$$
\dfrac{d}{d t}G(t) = (A+B+f'(t)H(A,B))G(t) \tag{4}
$$
Let $\dfrac{d}{d t}F(t) = \dfrac{d}{d t}G(t)$ :
\begin{align}
f'(t) = t \Rightarrow f(t) & = \dfrac{1}{2}t^2+C \tag{5} \\
H(A,B) & = [A,B] \tag{6} \\
F(t) & = G(t) \tag{7}
\end{align}
$$
(5)\&(6)\&(7) \quad \Rightarrow \quad e^{A t}e^{B t} = e^{A t+B t+(\dfrac{1}{2}t^2+C)[A,B]}
$$
- $t=0 \Rightarrow C=0$;
- $t=1 \Rightarrow $ $$\boxed{e^A e^B=e^{A+B+\dfrac{1}{2}[A,B]}}$$
Appendix for Hadamard's lemma:
$$
\boxed{e^{A t} B e^{-A t} = B+ t[A,B]+\dfrac{t^2}{2!}[A,[A,B]]+\dfrac{t^3}{3!}[A,[A,[A,B]]]+\cdots}
$$
Assume $Y(t) =e^{A t} B e^{-At} $
\begin{align}
Y^{(1)}(t) & = e^{At}(AB-BA)e^{-At} = e^{At}[A,B]e^{-At} \\
Y^{(2)}(t) & = e^{At} A [A,B]-[A,B] A e^{-At} = e^{At}[A,[A,B]]e^{-At} \\
Y^{(3)}(t) & = e^{At} [A,[A,[A,B]]] e^{-At} \\
& \cdots\cdots \nonumber
\end{align}
$$
\Rightarrow \quad Y(t) = \sum_{n=0}^{\infty} \dfrac{t^n}{n!}Y^{(n)}(t)|_{t=0}=B+t[A,B]+\dfrac{t^2}{2!}[A,[A,B]]+\dfrac{t^3}{3!}[A,[A,[A,B]]]+\cdots
$$