How proves the next formula?
Let $f,g\in S(R^n)$ Schwartz space. Then for all multi index $\alpha=(\alpha_1,\ldots, \alpha_n),\quad x^{\alpha}(f*g)(x)=\sum_{\gamma+\overline{\gamma}=\alpha}\frac{\alpha}{\gamma!\overline{\gamma}!} [(x^{\gamma}f)*(x^{\overline{\gamma}}g)](x)$.
I am trying to prove it by induction in the length of the multi index For $|\alpha|=1$, then $\alpha=(0,\ldots,0,\alpha_1,0,\ldots, 0)$ then is easy to see that formul holds.
Assuming the formula holds for $|\alpha|=k$. Let $\alpha'=(\alpha_1,\ldots, \alpha_n,\alpha_{n+1})$. Then \begin{align} x^{\alpha'}(f*g)(x)&=x_{n+1}^{\alpha_{n+1}} x^{\alpha}(f*g)(x)\\ &=x_{n+1}^{\alpha_{n+1}} \left(\sum_{\gamma+\overline{\gamma}=\alpha}\frac{\alpha}{\gamma!\overline{\gamma}!} [(x^{\gamma}f)*(x^{\overline{\gamma}}g)](x)\right)\\ &=\sum_{\gamma+\overline{\gamma}=\alpha}\frac{\alpha}{\gamma!\overline{\gamma}!} x_{n+1}^{\alpha_{n+1}} [(x^{\gamma}f)*(x^{\overline{\gamma}}g)](x)\\ &=\sum_{\gamma+\overline{\gamma}=\alpha}\frac{\alpha}{\gamma!\overline{\gamma}!} [(x_{n+1}^{\alpha_{n+1}}x^{\gamma}f)*(x^{\overline{\gamma}}g)+(x^{\gamma}f)*(x_{n+1}^{\alpha_{n+1}}x^{\overline{\gamma}}g)](x) \end{align} and here I can't think of how to reorder the indexes to leave everything as in the statement ... Actualization 1. I just saw that multiple indices can be any non-negative integer value. Given that, what I put up would not be correct and the problem would be even more complicated, because if $|\alpha|=1$ then $\alpha=(0,\ldots, 0, \alpha_1,0,\ldots, 0)$ but by example, If $\alpha_1=7$ then $\gamma+\overline{\gamma}\in \left\{0+7,7+0,1+6,6+1,2+5,5+2,3+4,4+3\right\}$ (abuse notation) complicating the formula much more.
Actualization 2.
\begin{align} &x^{\alpha}(f*g)(x)=(x-y+y)^{\alpha}(f*g)(x)\\ &=\sum_{0\leq \beta\leq \alpha} \frac{\alpha!}{\beta! (\alpha-\beta)!} (x-y)^{\beta}y^{\alpha-\beta}(f*g)(x)\\ &=\sum_{0\leq \beta\leq \alpha}\frac{\alpha!}{\beta! (\alpha-\beta)!} (x-y)^{\beta}y^{\alpha-\beta} \int f(x-y)g(y)dy\\ &=\sum_{0\leq \beta\leq \alpha}\frac{\alpha!}{\beta! (\alpha-\beta)!} [(x^{\beta}f)*(x^{\alpha-\beta}g)](x) \end{align}
I have already solved it :)