$e^{tA}e^{tB}=e^{t(A+B)}e^{\frac{t^2}{2}([A,B])}$ Proof that: $$e^{tB}e^{tA}=e^{t(A+B)}e^{\frac{t^2}{2}([A,B])}$$ for all $t\in \mathbb{R}$ if [A,[A,B]]=[B,[A,B]]=0
Hint: 


*

*$[A,B]=BA-AB$ and for hypothesis $A^2B+BA^2=2ABA$ and $AB^2+B^2A=2BAB$

*$\phi(t)=e^{-t(A+B)}e^{tB}e^{tA}$ is solution of $X'=t[A,B]X$
My progress:  $\phi'(t)=-(A+B)e^{-t(A+B)}e^{tB}e^{tA}+e^{-t(A+B)}Be^{tB}e^{tA}+e^{-t(A+B)}e^{tB}Ae^{tA}$
Then: $$\phi'(t)=[-(A+B) + e^{-t(A+B)}Be^{t(A+B)} + e^{-t(A+B)}e^{tB}Ae^{-tB}e^{-t(A+B)}]e^{-t(A+B)}e^{tB}e^{tA}$$
$$\phi'(t)=[-(A+B) + e^{-t(A+B)}Be^{t(A+B)} + e^{-t(A+B)}e^{tB}Ae^{-tB}e^{-t(A+B)}]\phi(t)$$
I should have $t(BA-AB)=[-(A+B) + e^{-t(A+B)}Be^{t(A+B)} + e^{-t(A+B)}e^{tB}Ae^{-tB}e^{-t(A+B)}]$ 
Thanks a lot.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets consider $\ds{\,\mrm{U}\pars{t} \equiv \expo{-At}\expo{\pars{A + B}t}}$.

Then,
\begin{align}
\partiald{\mrm{U}\pars{t}}{t} & =
\expo{-At}\pars{-A}\expo{\pars{A + B}t} +
\expo{-At}\pars{A + B}\expo{\pars{A + B}t} =
\expo{-At}B\expo{\pars{A + B}t}
\\[5mm] & =
\pars{\expo{-At}B\expo{At}}\pars{\expo{-At}\expo{\pars{A + B}t}}
\implies
\bbx{\partiald{\mrm{U}\pars{t}}{t} = \,\mrm{B}\pars{t}\,\mrm{U}\pars{t}}
\end{align}
where $\ds{\,\mrm{B}\pars{t} \equiv \expo{-At}B\expo{At}}$. Moreover,
$\ds{\,\mrm{U}\pars{t} = \mathbf{1} + \int_{0}^{t}\mrm{B}\pars{\tau}\,\mrm{U}\pars{\tau}\dd\tau}$. $\ds{\pars{~\mathbf{1}:\ identity~}}$. Note that
$$\bbx{%
\partiald{\mrm{B}\pars{t}}{t} = -A\,\mrm{B}\pars{t} + \,\mrm{B}\pars{t}A =
\bracks{\mrm{B}\pars{t},A}}
$$

$\Large\ds{\bracks{B,A} = 0.}$ In this case, $\ds{\mrm{B}\pars{t} = B \implies
\mrm{U}\pars{t} = \expo{Bt} \implies \expo{\pars{A + b}t} = \expo{At}\expo{Bt}}$

$\Large\ds{\bracks{B,A} = C \propto \mathbf{1} \implies
\bracks{\bracks{B,A},A} = 0.}$ Note that
\begin{align}
\partiald[2]{\mrm{B}\pars{t}}{t} & =
\bracks{\partiald{\mrm{B}\pars{t}}{t},A} =
\bracks{\bracks{\mrm{B}\pars{t},A},A} = 0
\\[5mm] \implies
\partiald{\mrm{B}\pars{t}}{t} & = \bracks{B,A}
\implies
\mrm{B}\pars{t} = B + \bracks{B,A}t 
\\[5mm] \implies
\mrm{U}\pars{t} & = \expo{Bt}\expo{\bracks{B,A}t^{2}/2}
\implies
\bbx{\expo{\pars{A + B}t} = \expo{At}\expo{Bt}\expo{\bracks{B,A}t^{2}/2}}  
\end{align}
A: As $B$ and $e^{Bt}$ commute, this is
$$\phi'(t)\phi(t)^{-1}
=-(A+B)+e^{-t(A+B)}e^{tB}(A+B)e^{-tB}e^{t(A+B)}$$
or
$$\phi'(t)\phi(t)^{-1}
=-C+e^{-tC}e^{tB}Ce^{-tB}e^{tC}$$
where $C=A+B$.
Call this $f(t)$.
Then
$$f'(t)=-Cf(t)+f(t)C+e^{-tC}e^{tB}(BC-CB)e^{-tB}e^{tC}.$$
Now $BC-CB=BA-AB$ commutes with $B$ and so $e^{tB}$.
Likewise it commutes with $e^{tC}$. Therefore
$$f'(t)=-Cf(t)+f(t)C+(BA-AB).$$
This looks promising; if only we could prove $Cf(t)=f(t)C\ldots$.
