$\Delta (u^m)=\mathrm{div}(m u^{m-1} \mathrm{grad}(u))$?

Question

How can one show that for a suitable $u$ the weak laplacian

$\Delta (u^m)$ equals $\mathrm{div}(m u^{m-1} \mathrm{grad}(u))$ both in the weak sense ($m>1$)

and what conditions need one to impose on $u$ then?