Inverse laplace transform of $1/s^4$

I know that the result of the inverse laplace transform of the following is

$$\mathcal{L}^{-1}\left\{ \frac{1}{s^4}\right\} = \frac{1}{6}\cdot t^3$$

However I just can't seem to figure out where the fraction comes from. $$\frac{1}{6}$$

Can anyone explain to me in short? I got the feeling that I'm missing out on something really stupid.. Thanks in advance.

$\mathcal{L}(t^3)=\frac{3!}{s^4}=\frac{6}{s^4}$ however we just have a $\frac{1}{s^4}$ so we need to get rid of that 6, to do this we divide by 6. So therefore $\mathcal{L}^{-1}\left(\frac{1}{s^4}\right )=\frac{t^3}{6}$
Hint. One may observe that by differentiating $$\int_0^\infty e^{-st}dt=\frac1s,\qquad s>0,$$ $n$ times one has $$\left(\int_0^\infty e^{-st}dt\right)^{(n)}=\int_0^\infty (-1)^n t^ne^{-st}dt,\qquad s>0,$$ and $$\left(\frac1s\right)^{(n)}=\left(s^{-1}\right)^{(n)}=(-1)^n \cdot n\cdot(n-1) \cdots 1 \cdot s^{-(n+1)}=(-1)^n\cdot n!\cdot s^{-(n+1)},\qquad s>0,$$ giving
$$\int_0^\infty \frac{t^n}{\color{red}{n!}}\:e^{-st}dt=\frac{1}{s^{\color{red}{n+1}}},\quad s>0,\quad n=0,1,2,\cdots.$$
The OP question corresponds to $s=1$, $n=3$.