 Fourier transform of a polynomial multiple

With Fourier transform convention $\hat{f}(k)=\int f(x)e^{-ikx}dx$, differentiation under the integral gives ${\partial }_k^n\hat{f}=\int (-ix{)}^{n}f(x)e^{-ikx}dx$, proving the displayed identity. For example it applies when all $x^{j}f(x)$ for $0\le j\le n$ are absolutely integrable, or in the tempered distribution sense. It is the dual counterpart of the Fourier transform of a derivative.