Bounds of higher order derivative of bandlimited signals
What simulations did you try? I think what you mention is right.
If $\mathcal{F}[f(t)] = F(\omega)$, according to the inverse Fourier transform, we have
$$
f(t) = \frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)e^{j\omega t}d\omega
$$
$$
f'(t) = \frac{d}{dt}\Big(\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)e^{j\omega t}d\omega\Big)= \frac{1}{2\pi}\int_{-\infty}^\infty j\omega F(\omega)e^{j\omega t}d\omega
$$
So the Fourier transform of derivative should be
$$\mathcal{F}[f'(t)] = j\omega F(\omega)$$
and the Fourier transform of $n$-th order derivative is
$$
\mathcal{F}[f^{(n)}(t)] = (j\omega)^nF(\omega)
$$