Proving an equation about exponential of matrices 
Let $A, B$ be $n\times n$ matrices over $\mathbb{C}$. Assume that
  $$[A,[A,B]]=[B,[A,B]] = 0$$
Show
  $$e^{\epsilon A}Be^{-\epsilon A} = B + \epsilon[A,B]$$

Then, I have to show the equation below. Here $[A, B]$ is equal to $AB-BA$ and $\epsilon$ is a real parameter. I can't find a way to show the equation below... Could anyone explain to me in detail?



$\textbf{Proof}$. Since 
$$
e^{\varepsilon A} = \sum_{k=0}^{\infty} \dfrac{(\varepsilon A)^k}{k!} 
= 1 + \varepsilon A + \frac{\varepsilon^2 A^2}{2!} + 
\frac{\varepsilon^3 A^3}{3!} + \ldots,  
$$ 
we have 
$$
e^{-\varepsilon A} = \sum_{k=0}^{\infty} \dfrac{(-\varepsilon A)^k}{k!} 
= 1 - \varepsilon A + \frac{\varepsilon^2 A^2}{2!} - 
\frac{\varepsilon^3 A^3}{3!} + \ldots.  
$$ 
So 
\begin{align*}
e^{\varepsilon A} B e^{-\varepsilon A} 
&= \left(1 + \varepsilon A + \frac{\varepsilon^2 A^2}{2!} + 
\frac{\varepsilon^3 A^3}{3!} + \ldots\right)B \left( 1 - \varepsilon A + \frac{\varepsilon^2 A^2}{2!} - 
\frac{\varepsilon^3 A^3}{3!} + \ldots\right) \\ 
&= \left(1 + \varepsilon A + \frac{\varepsilon^2 A^2}{2!} + 
\frac{\varepsilon^3 A^3}{3!} + \ldots\right) \left( B - \varepsilon B A + \frac{\varepsilon^2 B A^2}{2!} - 
\frac{\varepsilon^3 B A^3}{3!} + \ldots \right) \\ 
&= B + \varepsilon (AB-BA) + 2! \varepsilon^2 (A^2B-2ABA + BA^2) 
+3! \varepsilon^3[A,  [A,[A,B]] ]
+ \ldots \\ 
&= B + \varepsilon [A,B] + 2! \varepsilon^2 [A,[A,B]] +3! \varepsilon^3[A,  [A,[A,B]] ]
+ \ldots \\ 
&= B + \varepsilon [A,B] 
\end{align*}
since $[A,[A,B]]= A(AB-BA)-(AB-BA)A=A^2B-2ABA + BA^2=0$. $\hspace{2cm}\square$
