Is it correct to say that, if $\mathcal{L}\{f(t)\}=F(s)$, I have
$$\mathcal{L}^{-1}\{s^{n}F(s)\}=f^{(n)}(t)+c\delta(t),$$
where $c=\sum_{k=0}^{n-1} f^{(k)}(0)$?
I think as follows: $$\mathcal{L}\{f^{(n)}(t)+c\delta(t)\}=(s^{n}F(s)-c)+c\cdot 1=s^{n}F(s).$$
But I am not sure.
Thank you in advance.
The expression for $c$ in the OP is incorrect. In fact, $c$ depends on $s$, the transform variable. Specifically, we have
$$\begin{align} \mathcal{L} \{f^{(n)}\} &=s^n F(s)-\sum_{k=0}^{n-1}s^kf^{(n-1-k)}(0)\\\\ &=s^nF(s)-c(s) \end{align}$$
where $c(s)$ is given by
$$c(s)=\sum_{k=0}^{n-1}s^kf^{(n-1-k)}(0)$$
Therefore, the inverse Laplace transform of $s^nF(s)$ is
$$\begin{align} \mathscr{L}^{-1}\{s^nF(s)\}(t)&=f^{(n)}(t)+\sum_{k=0}^{n-1}f^{(n-1-k)}(0)\mathscr{L}^{-1}\{s^k\}(t)\\\\ &=f^{(n)}(t)+\sum_{k=0}^{n-1}f^{(n-1-k)}(0)\delta^{(k)}(t) \end{align}$$
where $\delta^{(k)}(t)$ is the $k$'th order distributional derivative of the Dirac Delta.