Proving a Simple Differential Equality by Induction I would like to prove the following statement.
For $\phi (r) \in C^{k+1}(\mathbb{R})$, and $k \in \mathbb{N}$:
$(d_{r}^{2}) (\frac{1}{r} d_{r})^{k-1} (r^{2k - 1} \phi(r)) = (\frac{1}{r}d_{r})^{k} (r^{2k}d_{r} \phi(r))$
Equality was easy to show for the $k=1$ case. I would then like to assume true for $k = n-1$ case and show that the $k=n$ case follows. However, I have been unable to do this. Could someone please help me with the induction step?
Here is my attempt:
Assume true for $k = n-1, \ n \in \mathbb{N}$. Consider the $k = n$ case.
We can rewrite the left-hand side of the equality as follows:
$(d_{r}^{2}) (\frac{1}{r} d_{r})^{n-1} (r^{2n - 1} \phi(r))$
$= (d_{r}^{2}) (\frac{1}{r} d_{r})^{k-2} (\frac{1}{r} d_{r} [r^{2k - 1} \phi(r)])$
$= (d_{r}^{2}) (\frac{1}{r} d_{r})^{k-2} (\frac{1}{r} [(2k - 1)r^{2k - 2} \phi(r) + r^{2k - 1} d_{r}\phi(r)])$
$= (d_{r}^{2}) (\frac{1}{r} d_{r})^{k-2} [(2k - 1)r^{2k - 3} \phi(r) + r^{2k - 2} d_{r}\phi(r)]$
Since the $k = n-1$ case is true, we can say:
$= (2k - 1) (\frac{1}{r} d_{r})^{k-1} (r^{2(k-1)} d_{r}\phi) + (d_{r}^{2}) (\frac{1}{r} d_{r})^{k-2} (r^{2(k-1)} d_{r} \phi)$
We now try to rewrite the right-hand side:
$(\frac{1}{r} d_{r})^{k} (r^{2k} d_{r} \phi) $
$= (\frac{1}{r} d_{r})^{k-1} (\frac{1}{r} [2kr^{2k-1} d_{r}\phi + r^{2k} d_{r}^{2} \phi ] )$
$= (\frac{1}{r} d_{r})^{k-1} (2kr^{2k-2} d_{r}\phi + r^{2k - 1} d_{r}^{2} \phi) $
Unfortunately, this is about as far as I can go! I have tried manipulating these in slightly different ways, but I just can't seem to get the desired result using the $k = n-1$ case. Could someone try to fix/point out a mistake I've made?
 A: 
$$(d_{r}^{2}) (\frac{1}{r} d_{r})^{n-1} (r^{2n - 1} \phi(r))
 = (d_{r}^{2}) (\frac{1}{r} d_{r})^{n-2} (\frac{1}{r} d_{r} [r^{2n - 1} \phi(r)])$$

The above equation is wrong. You can't take $\dfrac1r$ inside the derivative(why?). 
The first few steps are given below. Note that if $\phi\in C^{k+1}(\mathbb R)$, then $d_r\phi\in C^{k}(\mathbb R)$. Use this fact while applying induction. I hope you can take it from here:)
Suppose the result is true for $k=n-1$. Then for $k=n$,
\begin{align}
(d_{r}^{2}) \left(\frac{1}{r} d_{r}\right)^{n-1} (r^{2n - 1} \phi(r))
&=(d_{r}^{2}) \left(\frac{1}{r^{n-1}} d_{r}^{n-1} (r^{2n - 1} \phi(r)\right)\\
&=(d_{r}^{2}) \left(\frac{1}{r\cdot r^{n-2}} d_{r}^{n-2} (d_r(r^{2n - 1} \phi(r))\right)\\
&=(d_{r}^{2}) \left(\frac{1}{r\cdot r^{n-2}} d_{r}^{n-2} (r^{2n - 1} d_r\phi(r)+(2n-1)r^{2n-2}\phi(r))\right)\\
&=(d_{r}^{2}) \left(\frac{1}{r\cdot r^{n-2}} d_{r}^{n-2} (r^{2n - 1} d_r\phi(r))\right)\\&\qquad+(d_{r}^{2})\left(\frac{1}{r\cdot r^{n-2}}d_r^{n-2}(2n-1)r^{2n-2}\phi(r))\right)\\
&=(d_{r}^{2}) \left(\frac{1}{r\cdot r^{n-2}} d_{r}^{n-2} (r^{2n - 1} d_r\phi(r))\right)\\&\qquad+(2n-1)(d_{r}^{2})\left(\frac{1}{r\cdot r^{n-2}}d_r^{n-2}(r^{2n-2}\phi(r))\right).
\end{align}
