Frequency Differentiation Property of The Laplace Transform

If $ F(s) $ is the Laplace transform of $ f(t) $, then
$$F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t$$
Taking the derivative with respect to $ s $,
$$\frac{d F(s)}{d s}=\int_{0}^{\infty} f(t)\left(-t e^{-s t}\right) d t=\int_{0}^{\infty}(-t f(t)) e^{-s t} d t=\mathcal{L}[-t f(t)]$$
and the frequency differentiation property becomes
$$\mathcal{L}[t f(t)]=-\frac{d F(s)}{d s} \tag{1}$$
Repeated applications of this equation lead to
$$\mathcal{L}\left[t^{n} f(t)\right]=(-1)^{n} \frac{d^{n} F(s)}{d s^{n}}$$
For example, we know from Example 1 that $ \mathcal{L}\left[e^{-a t}\right]=1 / $ $ (s+a) $. Using the property in Eq. (1),
$$\mathcal{L}\left[t e^{-a t}\right]=-\frac{d}{d s}\left(\frac{1}{s+a}\right)=\frac{1}{(s+a)^{2}}$$
Note that if $ a=0 $, we obtain $ \mathcal{L}[t]=1 / s^{2} $ as in ramp function, and repeated applications of Eq. (1) will yield
$$\mathcal{L}\left[t^{n}\right]=\frac{n !}{s^{n+1}}$$