What is the commutator of the exponential derivative operator and the exponential position operator? \begin{align} \left[\exp(\partial_x),\exp(x)\right] =\exp(\partial_x)\exp(x) -\exp(x)\exp(\partial_x)=~? \end{align}
I first wrote down the exponentials into there power series sum \begin{align} \left[\exp(\partial_x),\exp(x)\right] = \sum_{n=0}^{\infty} \dfrac{\partial_x^n}{n!} \sum_{m=0}^{\infty} \dfrac{\partial_x^m}{m!} - \sum_{n=0}^{\infty} \dfrac{x^n}{n!} \sum_{m=0}^{\infty} \dfrac{\partial_x^m}{m!}, \end{align} and then grouped like terms \begin{align} \left[\exp(\partial_x),\exp(x)\right] = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \dfrac{ \left[ \partial_x^n x^m - x^n \partial_x^m \right] }{n!m!}, \end{align} and then evaluate the derivatives \begin{align} \left[\exp(\partial_x),\exp(x)\right] = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \dfrac{ \left[ \partial_x^n (x^m) + x^m \partial_x^n - x^n \partial_x^m \right] }{n!m!}. \end{align} I am not sure if$$ \partial_x^n\left(x^m\right)~=~\frac{m!}{\left(m-n+1\right)!} \, x^{m-n} \,,$$and then I got stuck.
