Inverse Laplace transform formula of $s^n$ using derivative of Laplace transform

Question

I know that the inverse laplace transform of $s^k$ is equal to $\delta^{(k)}$.

But, what about the following formula? I don't see it in any book or anything like that.

Using the fact that $L\{t^nf(t)\}=(-1)^n\cdot L\{f(t)\}^{(n)}$

Where $(n)$ denotes the $n-$th derivative.

Let $f(t) = L\{s^k\}(t)$

1. Applying $L^{-1}\{\}$ to both sides:

$f(t)=\frac{(-1)^n}{t^n}\cdot L^{-1}\{L\{s^k\}\}^{(n)}(t)$

2. Derivating $k$ times, i.e $n=k,$ i got that $L\{s^k\}^{(k)} = k!L\{1\}$

Finally $L\{s^k\}(t) = \frac{(-1)^k}{t^k}\cdot k!\cdot \delta(t)$

What is wrong with this development?(I think is wrong because I have not seen it in any book or class note)