Prove $\mathscr{F}[{x^nf(x)}] = (j)^n\times \mathscr{F}^n[f(x)]$

Question

How to prove this property of Fourier Transform:

$$\mathscr{F}[{x^nf(x)}] = (j)^n\times \mathscr{F}^n[f(x)]$$ Fourier Transform's definition is:
$$\mathscr{F}[f(x)] = \int_{-\infty}^{+\infty}{e^{-jwx}f(x)d(x)}$$

Answer 1

Consider the integral \begin{align} I_{1} = \int x \, e^{-i\omega x} \, f(x) \, dx. \end{align} This can be seen as \begin{align} I_{1} = i \frac{d}{d\omega} \int e^{-i \omega x} f(x) \, dx. \end{align} Extending this to \begin{align} I_{n} = \int x^{n} \, e^{-i \omega x} f(x) \, dx \end{align} leads to \begin{align} I_{n} = i^{n} \frac{d^{n}}{d\omega^{n}} \int e^{-i \omega x} f(x) \, dx. \end{align} Now it is seen that \begin{align} \mathscr{F}[{x^{n} \, f(x)}] = i^{n} \frac{d^{n}}{d\omega^{n}} \mathscr{F}[{f(x)}] \end{align} as required.