Confusing on an integral involving $(-\Delta)^{m}$

Question

I'm reading a paper on some class of evolution equations. I have a problem with the following integration: $$ \int_{\Omega}|v|^{r-2}v\ (-\Delta)^{m}v\ dx=(r-1)^{m}\int_{\Omega}|v|^{r-2}|D^{m}v|^{2}\ dx, \tag 1 $$ where $\Omega\subset R^{n}$, is open bdd and $2<r<\frac{2m}{n-2m}$ for $n>2m$ with $$ \frac{\partial^{i}v}{\partial\nu^{i}}=0,\quad for\quad i=0,1,\dots,m-1,\quad x\in\partial \Omega. $$ I recall that $D^{m}=\nabla .\nabla\dots\nabla$ and $D^{2m}=(\Delta)^{m}$. I'm confusing that how we can obtain the above integral by using Green's formula repeatedly. I think that above identity is true only for $m=1$. Since \begin{equation} \begin{aligned} \int_{\Omega}|v|^{r-2}v\ (-\Delta)^{m}v\ dx&=(-1)^{m+1}\int_{\Omega}D(|v|^{r-2}v)D^{2m-1}vdx\\&=(-1)^{m+2}\int_{\Omega}D^{2}(|v|^{r-2}v)D^{2m-2}vdx=\cdots=\int_{\Omega}D^{m}(|v|^{r-2}v)D^{m}vdx, \end{aligned} \end{equation} and this is not equal to the right hand side of (1). Am I right? If so, how I can derive an expression for $D^{m}(|v|^{q}v)$?