a proposition learnt from 2nd yr DE class says, Let $f$ have Laplca transform $F(s)$. Then $$ \mathcal{L}(t^n f(t))=(-1)^n\frac{d^n}{ds^n}F(s)$$
Proof.
Write $$ F(s)=\int^{\infty}_0 f(t)e^{-st}dt $$ Now differentiating under the integral sign gives \begin{align} \frac{d}{ds}F(s)&=\frac{d}{ds}\int^{\infty}_0 e^{-st}f(t)dt\\ &=\int^{\infty}_0 f(t)\frac{d}{ds} e^{-st}dt\\ &=-\int^{\infty}_0 t f(t) e^{-st} dt\\ &=-\mathcal{L}(t f(t)). \end{align} Thus $\mathcal{L}(t f(t))=-\frac{d}{ds}F(s)$. The general result now follows by induction.
Could anyone show me in detail how the general case follows by induction?
By your last identity,
$$\mathcal{L}(t^{n+1}f(t)) = \mathcal{L}(t\cdot t^{n}f(t))= -\frac{d}{ds}\mathcal{L}(t^{n}f(t)).$$
The induction hypothesis replaces this last with
$$=-\frac{d}{ds} \left( (-1)^n \frac{d^n}{ds^n} F(s) \right) = (-1)^{n+1} \frac{d^{n+1}}{ds^{n+1}} F(s). $$