Commutator of exponentiated operators $[e^\hat{A}, \hat{B}]$

Question

In Nakahara's derivation of the path-integral in "Geometry, Topology and Physics" the following identity is used $$ \partial_x^n e^{ikx} = e^{ikx}(ik + \partial_x)^n\tag{1} $$ to obtain $$ e^{-i\epsilon[-\partial_x^2 / 2m + V(x) ]} e^{ikx} = e^{ikx} e^{-i \epsilon [-(ik+\partial_x)^2/2m + V(x)]}\tag{2} $$

which seems to suggest you can just directly apply the first formula even though the operator $\partial_x^n$ is exponentiated. I want to understand how we get this second expression, is there some relation between $ [\hat{A}, \hat{B}]$ and $[\hat{A}, e^{\hat{B}}]$ which was used?

Answer 1

1. OP's two formulas are special cases of $$e^{A}f(B)e^{-A}~=~f(e^{A}Be^{-A})~=~f(e^{[A,\cdot]}B)~\stackrel{(2)}{=}~f(B+[A,B]).\tag{1}$$ The last equality in (1) holds if $$ [A,[A,B]]~=~0. \tag{2}$$

2. For a general relationship between $[e^{A},B]$ and $[A,B]$, see e.g. this Phys.SE post.