How to properly determine Green's Function for $\partial^n$? Given $G \in \mathcal{D}'(\mathbb{R})$, $n\in \mathbb{N}$ find the solution for   $$\partial_x^nG=\delta$$
I know that this solution should be $$G=\frac{x^{n-1}}{(n-1)!}\Theta(x)$$ but I am struggling to find a proper derivation.
I can prove that for $n=1$ one gets $G=\Theta(x)$ but after that my approaches start to get a bit too heretic.  
 A: You can use your proof for $n=1$ as a base case for an induction proof. So in the induction step you determine the action on a test function $\phi \in \mathcal{D}(\mathbb{R})$:
\begin{align}
\left\langle {\partial_x}^{n+1} \left(\frac{x^n}{n!} \Theta(x)\right), \phi\right\rangle &= (-1)^{n-1} \left\langle {\partial_x} \left(\frac{x^n}{n!} \Theta(x)\right), \partial_x \phi\right\rangle \quad \text{(derivative of distributions)} \\
&= (-1)^{n-1} \bigg[ \left\langle \frac{x^{n-1}}{(n-1)!} \Theta(x), {\partial_x}^{n-1}\phi\right\rangle \\
& \quad + \left\langle \frac{x^n}{n!}\delta, {\partial_x}^{n-1}\phi\right\rangle \bigg]~. \quad \text{(using product rule and base case)} \\
\end{align}
Now the first is exactly the action of ${\partial_x}^n(x^{n-1}/(n-1)!)$ on $\phi$ and thus equal to $\langle \delta, \phi \rangle$; the second term vanishes when you 'shift' the multiplication by $x^n/(n-1)!$ to the test function and apply the definition of the delta distribution:
$$
\left\langle \frac{x^n}{n!}\delta, {\partial_x}^{n-1}\phi\right\rangle = \left\langle \delta, \frac{x^n}{n!}{\partial_x}^{n-1}\phi\right\rangle = 0~.
$$
I'm guessing you can also approach the problem using Fourier transforms to derive the expression for $G$, which might seem more direct as you don't need to know the answer beforehand.
