Identity for Euler-Darboux-Poisson (Evans 2.4 lemma 2) The following identity is used in the treatment of the multivariate wave equation (Evans on p. 74)

Lemma 2 (Some useful identities). Let $\phi:\Bbb R\rightarrow \Bbb R$ be $C^{k+1}$. Then for $k=1,2,\dots$
$$\left(\frac{d^2}{dr^2}\right)\left(\frac 1r\frac{d}{dr}\right)^{k-1}\left(r^{2k-1}\phi(r)\right)=\left(\frac 1r\frac{d}{dr}\right)^{k}\left(r^{2k}\phi'(r)\right)$$

For $k=1$, both sides equal $2\phi'+r\phi''$; for $k=2$, they equal $8\phi'+7r\phi''+r^2\phi'''$. Abbreviating $\frac{d}{dr}$ with $d$, I think that the following identity will be useful in a proof by induction
\begin{aligned} d \frac 1r d &= \frac 1r d^{-1}d^2\\
&=\left(\frac 1r d + \frac 1{r^2}\right)d^{-1}d^2 \\
&=\frac 1r d^2 + \frac 1{r^2}d \end{aligned}
From this, it also follows that $d^2=rd\frac 1r d-\frac 1r d$. However, I tried to insert these identities in multiple places without success.
 A: I suppose $r\ne 0$ here. Fix $r_0\ne 0$. Let $P$ be the $(k+1)$th Taylor polynomial of $\phi$ at $r_0$, i.e., $P^{(j)}(r_0) = \phi^{(j)}(r_0)$ for $j=0, \dots, k+1$. Note that if we expand both sides of the claimed identity, they will have derivatives of $\phi$ only up to order $k+1$. Therefore, both sides, evaluated at $r_0$, remain the same if we replace $\phi$ by $P$. 
So if suffices to prove the identity for polynomials. For that, it suffices to prove it for monomials (by linearity). For a monomial $r^n$, the statement becomes
$$\left(\frac{d^2}{dr^2}\right)\left(\frac 1r\frac{d}{dr}\right)^{k-1}\left(r^{2k+n-1}\right) = 
 n\left(\frac 1r\frac{d}{dr}\right)^{k}\left(r^{2k+n-1}\right)\tag1$$
Let $$Q(r) = \left(\frac{d}{dr}\right)\left(\frac 1r\frac{d}{dr}\right)^{k-1}\left(r^{2k+n-1}\right)$$
This is a monomial of degree $2k+n-1 - 2(k-1) -1 =n$. That is,    $Q$ is $cr^n$ for some constant $c$. In terms of $Q$, the identity (1) becomes
$$
\frac{d}{dr}Q  = n\frac{1}{r}Q
$$
which is obvious: both sides are $ncr^{n-1}$.
A: Suppose the identity holds for $k-1$, the for k:
LHS=$\frac{d^2}{dr^2}(\frac{1}{r}\frac{d}{dr})^{k-1}(r^{2k-1}\phi)=\frac{d^2}{dr^2}(\frac{1}{r}\frac{d}{dr})^{k-2}((2k-1)r^{2k-3}\phi+r^{2k-2}\phi')$
RHS=$(\frac{1}{r}\frac{d}{dr})^{k}(r^{2k}\phi')=(\frac{1}{r}\frac{d}{dr})^{k-1}(2kr^{2k-2}\phi'+r^{2k-1}\phi'')$
By induction hypothesis, the first term of RHS can be merged with the first term of LHS.
LHS-RHS=$\frac{d^2}{dr^2}(\frac{1}{r}\frac{d}{dr})^{k-1}(r^{2k-3}(r\phi'-\phi))-(\frac{1}{r}\frac{d}{dr})^{k-1}(r^{2k-1}\phi'')$
Use induction hypothesis again with $(r\phi'-\phi)$ replacing $\phi$. 
=$(\frac{1}{r}\frac{d}{dr})^{k-1}(r^{2k-2}(r\phi'-\phi)')-(\frac{1}{r}\frac{d}{dr})^{k-1}(r^{2k-1}\phi'')=0$
A: consider$$ \left(\frac{1}{r} \frac{d}{d r}\right)(t^m\phi)=r^{n-2}\left(r\frac{d}{d r}+m\right)\phi$$
so$$\left(\frac{1}{r} \frac{d}{d r}\right)^{k-1}\left(r^{2 k-1} \phi\right)=r\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\phi$$
change $\phi$ into $r\frac{d}{dr}\phi$ ,we get$$\left(\frac{1}{r} \frac{d}{d r}\right)^{k-1}\left(r^{2 k} \frac{d \phi}{d r}\right)=r\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\left(r\frac{d}{d r}\right)\phi$$
notice that$ \frac{d}{dr}r\varphi=\left(r\frac{d}{d r}+1\right)\varphi$,so
$$
\begin{align*}
\left(\frac{d}{dr}\right)^2\left(\frac{1}{r} \frac{d}{d r}\right)^{k-1}\left(r^{2 k-1} \phi\right)&=\left(\frac{d}{dr}\right)^2r\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\phi\\
&=\left(\frac{d}{dr}\right)\left(r\frac{d}{d r}+1\right)\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\phi
\end{align*}
$$
and$$\begin{align*}
\left(\frac{1}{r} \frac{d}{d r}\right)^{k-1}\left(r^{2 k} \frac{d \phi}{d r}\right)&=\frac{1}{r}\left(r\frac{d}{d r}+1\right)\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\left(r\frac{d}{d r}\right)\phi\\
&=\frac{1}{r}\left(r\frac{d}{d r}\right)\left(r\frac{d}{d r}+1\right)\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\phi\\
&=\left(\frac{d}{d r}\right)\left(r\frac{d}{d r}+1\right)\left(r\frac{d}{d r}+3\right)\left(r\frac{d}{d r}+5\right)\cdots\left(r\frac{d}{d r}+2k-1\right)\phi
\end{align*} $$
