Why $$(2\pi x)^\alpha f(x)(2\pi i\xi)^\beta e^{-2i\pi x\cdot \xi}=(\partial _x)^\beta [(2\pi x)^\alpha f(x)]e^{-2i\pi x\cdot \xi} ?$$
Indeed, to prove that the fourier transform is in the Schwarz space, we do as following \begin{align*} (2\pi i\xi)^\beta (i\partial _\xi)^\alpha \hat f(\xi)&=\int_{\mathbb R^n}(2\pi x)^\alpha f(x)(2i\pi i\xi)^\beta e^{-2\pi ix\cdot \xi}dx\\ &=\int_{\mathbb R^n}(2\pi x)^\alpha f(x)(-\partial _x)^\beta e^{-2\pi ix\cdot \xi}dx\\ &\underset{(*)}{=}\int_{\mathbb R^n}(\partial _x)^\beta [(2\pi x)^\alpha f(x)]e^{-2i\pi x\cdot \xi}dx \end{align*} but I don't understand the equality $(*)$. I recall that $\alpha,\beta \in \mathbb N^n$ and that we used $$x^\alpha =\prod_{i=1}^n x_i^{\alpha _i}.$$