Differential identity $\left(x^2\frac{d}{dx}\right)^nf(x)=x^{n+1}\frac{d^n}{dx^n}\left(x^{n-1}f(x)\right)$ I have found the following differential identity:
$$\left(-x^2\frac{d}{dx}\right)^nf(x)=(-1)^n x^{n+1}\frac{d^n}{dx^n}\left(x^{n-1}f(x)\right)$$
I have used it to find an alternative Rodrigues representation for Bessel polynomials.
My proof
we have placed: $D=\frac{d}{dx}$ in order to save space.
We start observing that, for n=1 the following is trivial that:
\begin{equation}
\left(x^{2}D\right)=x^{n+1}D^{n}x^{n-1},n=1
\end{equation}
Therefore for n=1 the identity holds. For induction we proof that,
if the identity holds for n, then it holds also for n+1. Applying
the operator$\left(x^{2}D\right)$ 
to both members:
\begin{equation}
\left(x^{2}D\right)\left(x^{2}D\right)^{n}=\left(x^{2}D\right)\left(x^{n+1}D^{n}x^{n-1}\right)
\end{equation}
\begin{equation}
\left(x^{2}D\right)^{n+1}=x^{2}Dxx^{n}D^{n}x^{n-1}=x^{2}Dx^{n}xD^{n}x^{n-1}
\end{equation}
remembering the following commutation rules:
\begin{equation}
Dx^{n}\centerdot-x^{n}D=nx^{n-1}
\end{equation}
\begin{equation}
D^{n}x\centerdot-xD^{n}=nD^{n-1}
\end{equation}
we can write:
\begin{equation}
\left(x^{2}D\right)\left(x^{2}D\right)^{n}=x^{2}\left[Dx^{n}\centerdot\right]\left[xD^{n}\right]x^{n-1}=x^{2}\left[x^{n}D+nx^{n-1}\right]\left[D^{n}x\centerdot-nD^{n-1}\right]x^{n-1}
\end{equation}
\begin{equation}
=x^{n+2}D^{n+1}x^{n}\centerdot+x^{2}\left[nx^{n-1}D^{n}x\centerdot-x^{n}DnD^{n-1}-nx^{n-1}nD^{n-1}\right]x^{n-1}
\end{equation}
Now we can show that the term in square parentheses is null, indeed
applying above seen commutation rules:
\begin{equation}
=x^{n+2}D^{n+1}x^{n}\centerdot+x^{2}\left[nx^{n-1}xD^{n}+nx^{n-1}nD^{n-1}-nx^{n}D{}^{n}-nx^{n-1}nD^{n-1}\right]x^{n-1}=
\end{equation}
\begin{equation}
=x^{n+2}D^{n+1}x^{n}\centerdot+x^{2}\left[nx^{n}D^{n}-nx^{n}D{}^{n}+nx^{n-1}nD^{n-1}-nx^{n-1}nD^{n-1}\right]x^{n-1}=x^{n+2}D^{n+1}x^{n}\centerdot
\end{equation}
Therefore:
\begin{equation}
\left(x^{2}D\right)^{n+1}=\left(x^{2}D\right)\left(x^{2}D\right)^{n}=x^{n+2}D^{n+1}x^{n}\centerdot
\end{equation}
Questions
Is my proof correct?
Is this identity already known? If yes, what are the references?
Where can I find a possibly more elegant proof? 
Edit
So, it seems the identity was true, and already known.
In "Mathematical Analysis I" (Springer), Zorich V.A, in chapter "The Basic Rules of Differentiation", exercise 3 at page 213, the identity of $D^nf(1/x)=(-1)^nx^{n+1}D^n x^{n-1}f(1/x)$ has to be demonstrated. Unfortunately no hint was given and no proof is sketched, so I remain without a more straightforward proof ...) 

I think you can write your identity better as your $\frac{d}{d(\frac{1}{x})}$ is undefined. You can write your identity like that:
$\frac{1}{x^{n+1}}f^{(n)}(\frac{1}{x})=(-1)^n\frac{d^n}{dx^n}(x^{n-1}f(\frac{1}{x}))$
To prove it, you can do this, without using the differentiation operator (I hate using it...):
1- The base case where $n=0$ is trivial. Prove that and the case where $n=1$. This should be very easy.
2- Now, start from the right side like this:
$(-1)^{n+1}\frac{d^{n+1}}{dx^{n+1}}(x^{n}f(\frac{1}{x}))=(-1)^{n+1}(\frac{d^n}{dx^n}(x\cdot x^{n-1}f(\frac{1}{x})))'$
Use the General Leibniz Rule and the product, knowing that any $n$th derivative of $x$ where $n\geq2$ is 0 and use the base case you proved, to follow by induction.
QED.
Also, as far as I know, it doesn't have a name. I have seen this in Zorich, and seeing that he didn't mention a name for it though he usually names important theorems in exercises, I believe it doesn't have one.
