Is there a direct way to get $\widehat{f^{(n)}}(\omega)=(i \omega)^{n} \hat{f}(\omega)$ without induction on $n$ ?
I tried to use integration by parts for $$\mathcal{F}\{ f_n(x) \}=\widehat{f^{(n)}}(\omega)=\int_{-\infty}^{\infty}f^{(n)}(x) e^{-i \omega x} dx,$$ with $u=e^{-i \omega x}$ and $v=f^{(n+1)}(x)/{(n+1)}.$ Any suggestion is much appreciated.
$$ \begin{align*} \frac{\mathrm d^n}{\mathrm dx^n}f(x)&=\frac{\mathrm d^n}{\mathrm dx^n}\left(\frac{1}{2\pi}\int_{-\infty}^\infty \hat f(\omega)\,\mathrm e^{i\omega x}\,\mathrm d\omega\right)\\ &=\frac{1}{2\pi}\int_{-\infty}^\infty \hat f(\omega)\frac{\mathrm d^n}{\mathrm dx^n}\,\left(\mathrm e^{i\omega x}\right)\,\mathrm d\omega\\ &=\frac{1}{2\pi}\int_{-\infty}^\infty(i\omega)^n\hat f(\omega)\,\mathrm e^{i\omega x}\,\mathrm d\omega\\ &=\mathcal F^{-1}\{(i\omega)^n\hat f(\omega)\} \end{align*} $$ and then $\widehat{f^{(n)}}(\omega)=(i \omega)^{n} \hat{f}(\omega)$