I am trying to prove the following
Let $F$, $\Phi$ and $\phi$ be arbitrarily differentiable functions of $x$. I have come across the following identity in some literature I am reading which I need help in proving:
$\frac{1}{\phi^{r-1}}D^{r-1}\left[\frac{(F-\Phi)^r}{\phi}\right]=D_{(r-1)}\left[\left(\frac{F-\Phi}{\phi}\right)^{r}\right],\qquad r=1,2,\ldots$
where $\displaystyle D^r=\frac{d^r}{dx^r}$ and D_{(r)}=\left(D-\frac{\phi'}{\phi}\right)\left(D-2\frac{\phi'}{\phi}\right)\cdots\left(D-r\frac{\phi'}{\phi}\right) and $D_0$ is just the identity.
I can see this for the case $r=2$: \begin{align*} \frac{1}{\phi}D\left[\frac{(F-\Phi)^{2}}{\phi}\right]&=\frac{1}{\phi}\left[\frac{\phi D(F-\Phi)^{2}-(F-\Phi)^{2}\phi'}{\phi^{2}}\right]\\ &=\left(D-\frac{\phi'}{\phi}\right)\left[\frac{(F-\Phi)^{2}}{\phi^{2}}\right]\\ &=D_{(1)}\left[\left(\frac{F-\Phi}{\phi}\right)^{2}\right]\end{align*}
Its just an application of the quotient rule and some algebra. But how would I prove this for an arbitrary positive integer $r$?