special case of Leibniz rule $f,g : \mathbb{R} \rightarrow \mathbb{R} $ are n times differentiable.
I would like to proof that
$f*\frac{\mathrm{d}^{n} }{\mathrm{d} x^{n}}(g) = \sum_{k=0}^{n}((-1)^{n} \binom{n}{k} \frac{\mathrm{d}^{n-k} }{\mathrm{d} x^{n-k}}(f^{(k)}*g))$
I made a few examples but I have no idea how to proof. I tried with induction but failed. Can't find it anywhere else in the internet. I'm pretty sure it has something to do with the general Leibniz rule but can't figure out in which way.
Does anyone have an hint for me?
 A: We  use  the general Leibniz rule to show this identity. 

Writing $D$ for $\frac{d}{dx}$ we obtain
  \begin{align*}
\sum_{k=0}^n&(-1)^k\binom{n}{k}D^{n-k}\left(\left(D^kf\right)\cdot g\right)\\
&=\sum_{k=0}^n(-1)^k\binom{n}{k}\sum_{j=0}^{n-k}\binom{n-k}{j}\left(D^{n-j}f\right)\cdot\left(D^jg\right)\tag{1}\\
&=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\sum_{j=0}^{k}\binom{k}{j}\left(D^{n-j}f\right)\cdot\left(D^jg\right)\tag{2}\\
&=\sum_{j=0}^{n}\sum_{k=j}^n(-1)^{n-k}\binom{n}{j}\binom{n-j}{k-j}\left(D^{n-j}f\right)\cdot\left(D^jg\right)\tag{3}\\
&=\sum_{j=0}^{n}\binom{n}{j}\left(D^{n-j}f\right)\cdot\left(D^jg\right)\sum_{k=0}^{n-j}(-1)^{n-k-j}\binom{n-j}{k}\tag{4}\\
&=\sum_{j=0}^{n}\binom{n}{j}\left(D^{n-j}f\right)\cdot\left(D^{j}g\right)\sum_{k=0}^{n-j}(-1)^{k}\binom{n-j}{k}\tag{5}\\
&=\sum_{j=0}^{n}\binom{n}{j}\left(D^{n-j}f\right)\cdot\left(D^{j}g\right)(1-1)^{n-j}\tag{6}\\
&=f\cdot  (D^n g)
\end{align*}
  and the claim follows.

Comment:


*

*In (1)  we apply the general Leibniz rule.

*In (2) we revert the summation order of the outer sum by setting $k\longrightarrow n-k$.

*In (3) we exchange the sums by noting that
$$\sum_{k=0}^n \sum_{j=0}^{k} a_{k,j}=\sum_{0\leq j\leq k\leq n} a_{k,j}=\sum_{j=0}^n\sum_{k=j}^n a_{k,j}$$
and we also use the binomial identity $\binom{n}{k}\binom{k}{j}=\binom{n}{j}\binom{n-j}{k-j}$.

*In (4) we change the index of the inner to sum to start with $k=0$.

*In (5) we revert the summation order of the inner sum by setting $k\longrightarrow n-j-k$.

*In (6) we use the binomial formula and observe that only $n=j$ gives a non-zero contribution.
