Let $ A, B \in \mathbb {R}^{n \times n} $, the Lie bracket of $ A $ and $ B $ denoted by $ [A, B] $ is defined as $ [A, B] = BA-AB$. If $\big[A,[A, B] \big] = \big[B, [A, B] \big] = 0$, provide that for all $ t\in\mathbb{R}$ it is fulfilled that , $$e^{tB}e^{tA} = e^{t(A + B)}e^{t^{2}/2[A, B]} $$
I am given by suggestion, consider that $ \phi (t) = e ^ {- t (A + B)} e ^ {tB} e ^ {tA} $ is solution of the matrix equation $ X '= t [A, B] X $
So compute $$ ϕ'(t)=e^{-t(A+B)}\Bigl[e^{tA}B+(A-(A+B))e^{tA}\Bigr]e^{tB} =e^{-t(A+B)}\Bigl[e^{tA}B-Be^{tA}\Bigr]e^{tB} $$ Now consider $$ e^{tA}B-Be^{tA}=\sum_k\frac{t^k}{k!}(A^kB-BA^k)=\sum_k\frac{t^k}{k!}\,\sum_{m=0}^{k-1}A^m[A,B]A^{k-1-m}=[A,B]e^{tA} $$ the last equality because of the assumption that $A$ and $[A,B]$ commute. So we got $$ ϕ'(t)=e^{-t(A+B)}[A,B]e^{tA}e^{tB} $$ and because also $B$ commutes with $[A,B]$, so does $(A+B)$ resulting in $$ ϕ'(t)=[A,B]ϕ(t) $$ as claimed. Now also observe $ϕ(0)=I$ so that $ϕ(t)=e^{t[A,B]}$.