Let $f \in C^k[0,+\infty)$ and $ \sigma(f):=\inf\{s \in \mathbb{R}:\exp(-st)f(t) \in L^1[0,+\infty)\}$.
I want to show that for $\Re z>\max_{j=0,\ldots,k}\sigma(f^{(j)})$
$$\mathcal L(f^{(k)})(z)=z^k \mathcal L (f)(z)-\sum_{j=0}^{k-1}f^{(j)}(0)z^{k-1-j}$$
I tried the following, using induction and partial integration. I used the induction hypothesis in the third step.
Can this be done like this? And does $e^{-zt}f^{(k-1)}(t)$ equal $0$ when plugging for $t=\infty$?
Well, first of all if we define $$ F(z) = \mathcal{L}(f)(z) $$ The basic case for $k = 0$ essentially follows from the definition above. Now if you assume $$ \mathcal L(f^{(k-1)})(z)=z^{k-1} F(z) -\sum_{j=0}^{k-2}f^{(j)}(0)z^{k-1-j} $$ you can use integration by parts for case $k$ since $$ \mathcal{L}(f^{(k)})(z) = \mathcal{L}(Df^{(k-1)})(z) = \left. f^{(k-1)}(t)e^{-zt} \right|_{0}^{+\infty} + z \mathcal{L}(f^{(k-1)})(z) = -f^{(k-1)}(0) + z \mathcal{L}(f^{(k-1)})(z), $$ here $D$ denotes the differential operator. If you substitute to $\mathcal{L}(f^{(k-1)})(z)$ the expression given by your inductive hypothesis you'll get your results.
Of course to make the integrals valid you need to use your other assumptions, I just focused on the calculations.