Bounds of higher order derivative of bandlimited signals Here is a bandlimited function f(t) with bandwidth Ω:

The function f(t) is bounded in [-A,A].
Then the bound of the derivative of f(t) is bounded as:
|f'(t)|≤2πΩA.
So, what is the bound of its n-th order derivatives?
Is it as follows?

where n is the order of derivative.
If yes, then are the bounds of higher order derivatives much larger than the bound of f(t)?
I tried MATLAB simulations but the results were contrary.
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
$$
\mathcal{F}[f^{(n)}(t)] = (j\omega)^nF(\omega)
$$