If $[A, [A, B]] = 0 = [B, [A, B]]$, where $[A,B]=AB-BA$, then
$$e^{tA} e^{tB}=e^{t(A+B)+(t^2/2)[A,B]}$$
The book suggests proving that $e^{t(A+B)+(t^2/2)[A,B]}$ is solution of $$\dot{X}=AX+XB, \qquad X(0)=I$$ if and only if $$\frac{d}{dt}( e^{-(t(A+B)+(t^2/2)[A,B])}Ae^{t(A+B)+(t^2/2)[A,B]} - t[A,B])=0$$
I derived the function $t \mapsto e^{t(A+B)+(t^2/2)[A,B]}$, but I couldn't show that is a solution of the initial value problem or use the hint.
This is exercise 2.52 in the book of Differential Equations from Chicone.
One can show by induction that for all $n\in\mathbb{N}^*$, $$[A,B^n]=nB^{n-1}[A,B]$$ Thus $$ \forall n\in\mathbb{N}^*, A\frac{B^n}{n!}-\frac{B^n}{n!}A=\frac{B^{n-1}}{(n-1)!}[A,B] $$ Suming this gives $$ [A,\exp(B)]=\exp(B)[A,B] $$ Let $\varphi(t)=\exp\left(t(A+B)+\frac{t^2}{2}[A,B]\right)$ then $\varphi'(t)=(A+B+t[A,B])\varphi(t)$ and $\varphi(0)=I_n$. Now let $\psi(t)=\exp(tA)\exp(tB)$, then $$ \psi'(t)=\exp(tA)(A\exp(tB)+\exp(tB)B) $$ and $[tA,\exp(tB)]=\exp(tB)[tA,tB]$ that is to say $A\exp(tB)-\exp(tB)A=t\exp(tB)[A,B]$ so we have $$ \psi'(t)=(A+B+t[A,B])\psi'(t) $$ Since $\psi(0)=I_n$, according to Cauchy-Lipschitz theorem we have $\psi=\varphi$ that is to say $$ \exp(tA)\exp(tB)=\exp\left(tA+tB+\frac{t^2}{2}[A,B]\right) $$