= Solution
For an integer $n\ge0$ and $\operatorname{Re}s>0$, the definition of the <Laplace transform> gives $F_0(s)=\int_0^\infty e^{-st}\,dt=1/s$. For $n\ge1$, <integration by parts> gives
$$
F_n(s)=\left[-\frac{t^ne^{-st}}s\right]_0^\infty+\frac ns\int_0^\infty t^{n-1}e^{-st}\,dt=\frac nsF_{n-1}(s).
$$
The endpoint term vanishes because <exponential decay> dominates every fixed power of $t$. Induction therefore gives $F_n(s)=n!/s^{n+1}$. Multiplication by $e^{at}$ replaces $e^{-st}$ by $e^{-(s-a)t}$ in the defining <integral>, so
$$
\boxed{\mathcal L(t^n)(s)=\frac{n!}{s^{n+1}},\qquad\mathcal L(e^{at}t^n)(s)=\frac{n!}{(s-a)^{n+1}}.}
$$
The second formula holds for $\operatorname{Re}(s-a)>0$ and is the <Laplace transform shift theorem>.
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