I need some help to compute the k-th derivative of the Dirac's Delta function, $\delta_0^{(k)}$.
I know its Fourier transform is $TF(\delta_0^{(k)})(y)=(iy)^{k}$( I don't know if this could be useful).
Thanks a lot for any help.
EDIT. Context of the question:
I'm trying to obtain the fourier transform of $y^k$. To do that, I am trying to use the next expression, that I have proved: $TF(\delta_{0}^{(k)})(y)=(iy)^k$.
Then, if I apply Fourier transform to the previous expression, I have:
$\delta_0^{(k)}(-y)=TF(\small{TF(\delta_{0}^{(k)})})(y)=i^k\;TF(y^k)$
and this implies $\delta_0^{(k)}(y)=i^k\;TF((-y)^k)$, which leads to $TF(y^k)=\frac{1}{(-1)^ki^k}\overbrace{\delta_0^{(k)}(y)}^{*}$,
And that's why I need to obtain an expression for *.
The comment contains the correct answer. Distributions are only defined in terms of how they act when integrated against test functions. So with an abuse of notation: $$ \int_{\mathbb R} \phi(x) \delta(x) \, dx = \phi(0) ,$$ and by integrating by parts $k$ times $$ \int_{\mathbb R} \phi(x) \delta^{(k)}(x) \, dx = (-1)^{k-1} \phi^{(k)}(0) .$$
You can also see this intuitively by approximating the $\delta$ function by $n \psi(n x)$, where $\psi(x)$ is smooth, non-negative, supported in a neighbourhood of $0$, and $\int_{\mathbb R} \psi(x) \, dx = 1$, and then letting $n$ be very large. Think about what the derivatives of this might look like, and you will see why the second displayed equation is intuitively reasonable.