I write down the Taylor (Mclaurin) expansion of the analytic function f: \begin{equation} f(x) = \sum_{k} \frac{x^k}{k!}\frac{d^k}{dx^k}\bigg|_{x=0} (f(x)) \end{equation}
It is easy to show that \begin{equation} \hat{A}=x\frac{d}{dx} \end{equation} is linear, therefore also \begin{equation} \hat{A^k}=x^k\frac{d^k}{dx^k} \end{equation} and \begin{equation} \sum_{k} \frac{x^k}{k!}\frac{d^k}{dx^k}\bigg|_{x=0} \end{equation} are linear too. I then conclude that \begin{equation} f(x) = \lbrace\sum_{k} \frac{x^k}{k!}\frac{d^k}{dx^k}\bigg|_{x=0}\rbrace f(x)\\ f(x) = exp (\hat{A}) f(x)\\ exp (\hat{A}) = exp (x\frac{d}{dx}) = 1 = exp (0) = exp (2\pi i) \end{equation}
I consider now the equation \begin{equation} x\frac{d}{dx} = 2\pi i \end{equation} is it totally wrong? why? are there some weaker assumption that could be made in order to make sense to the whole thing?