Suppose x is a real number, and $\hat{A}$ and $\hat{B}$ are two non-commuting operators.
A polynomial operator is defined by
$$g(\hat{B}) = \sum_{n=0}^\infty a_n \hat{B}^n,$$ where $a_n$ are real numbers.
If [$\hat{A}$ , $\hat{B}$] = c, where c is a complex number, prove that
$$exp(x\hat{A})g(\hat{B})exp(-x\hat{A}) = g(\hat{B} + cx).$$
A previous proof yielded the taylor expansion of a similar expression, which led me to believe it could be the key, thus i started at
$$exp(x\hat{A})g(\hat{B})exp(-x\hat{A}) = \sum_{n=0}^\infty [\hat{A},g(\hat{B})]_n \frac{x^n}{n!}.$$
Here, $[\hat{A} , \hat{B}]_n$ := $[\hat{A},[\hat{A},\hat{B}]]_{n-1}$, describes the nested commutator, with $[\hat{A},\hat{B}]_0 := \hat{B}.$
Sadly, I could not make it work as of now. Am I missing an obvious or simple solution? Any help is greatly appreciated.
You are essentially there!
Observe $\hat A/c$ acts like a derivative operator on powers of $\hat B$, and so $g(\hat B)$ as well, $$ [\hat A, g(\hat B)]= c g^{(1)}(\hat B), \qquad \Longrightarrow \\ [\hat A, g(\hat B)]_n= c^n g^{(n)} (\hat B), $$ where the superscript in parenthesis denotes number of derivatives; so that, substituting in your final (Duhamel) expression readily yields the Taylor expansion at $\hat B$, $$\exp(x\hat{A})g(\hat{B})\exp(-x\hat{A}) = \sum_{n=0}^\infty [\hat{A},g(\hat{B})]_n ~ \frac{x^n}{n!}\\ = \sum_{n=0}^\infty g^{(n)} (\hat B) ~\frac{c^n x^n}{n!} = g(\hat B + cx) ~ .$$
(Historically, this was Baker's minority contribution to the CBH expansion algorithm.)