Proof of the Laplace transform of the $n$th derivative

You've got the basis case $n=1$. If $g=f^{(k)}$ and $$\mathcal{L}(g) = \mathcal{L}(f^{(k)}) = s^k \mathcal{L}(f) - s^{k-1}f(0) - \dotsb - f^{(k-1)}(0), $$ then using the $n=1$ result gives \begin{align} \mathcal{L}(f^{(k+1)})=\mathcal{L}(g') &= s\mathcal{L}(g) - g(0) \\ &= s(s^k \mathcal{L}(f) - s^{k-1}f(0) - \dotsb - f^{(k-1)}(0)) - f^{(k)}(0) \\ &= s^{k+1} \mathcal{L}(f) - s^{k}f(0) - \dotsb - sf^{(k-1)}(0)) - f^{(k)}(0), \end{align} which is of the same form with $k \mapsto k+1$, so combining this with the basis case, induction implies the result holds for all positive integers $n$.