Prove that if $f(x_1,...,x_n)$ is homogeneous of degree $p$, i.e; $f(tx)=t^pf(x)$. Then:
$$(x_1 \frac { \partial}{\partial x_1} +...+x_n \frac { \partial}{\partial x_n})^mf(x_1,...,x_n)=p(p-1)(p-2)...(p-m+1)f(x_1,...,x_n)$$
I've tried using the multinomial theorem and doing some induction over $m$, however I haven't been able to prove it even for $m=2$. Is this approach right, or what would you do?
Be $y=tx$. Then on one hand, you have $$\left.\frac{\mathrm d}{\mathrm dt} f(y)\right|_{t=1} = \left.\frac{\mathrm d}{\mathrm dt} f(tx)\right|_{t=1} = \left.\frac{\mathrm d}{\mathrm dt} t^p f(x)\right|_{t=1} = \left.\vphantom{\frac{\mathrm d}{\mathrm dt}} p t^{p-1} f(x)\right|_{t=1} = p f(x)$$ On the other hand, you have $$\left.\frac{\mathrm d}{\mathrm dt} f(y)\right|_{t=1} = \left.\left(\frac{\mathrm dy_1}{\mathrm dt}\frac{\partial}{\partial y_1} + \ldots + \frac{\mathrm dy_n}{\mathrm dt}\frac{\partial}{\partial y_n}\right)f(y_1,\ldots,y_n)\right|_{t=1} = \left(x_1 \frac{\partial}{\partial x_1} + \ldots + x_n \frac{\partial}{\partial x_n}\right)f(x)$$
It should be obvious how to extend that for higher derivatives.