Relation involving generalized Laguerre polynomials Playing around with different approaches to solve the radial part of the Schrodinger equation for the hydrogen-like atom, I have obtained the following expression ($l$ and $n$ are non-negative integers)
$$ \tilde L^{(2l+1)}_n(x) = \frac{1}{n!} x^{-(l+1)}e^{x/2} \left(x \frac{d}{dx}-x \frac{d^2}{dx^2}  + \frac{l(l+1)}{x} -\frac{x}{4} \right)^n \bigl(x^{l+1} e^{-x/2} \bigr)
$$
which I believe is a
representation of the generalized Laguerre polynomials.
The latter are conventionally defined as via de Rodrigues formula
$$L_n^{(2l+1)}(x) = \frac{1}{n!} x^{-(2l+1)} e^{x} \frac{d^n}{dx^n} \Bigl(x^{2l+1+n} e^{-x} \Bigr)\,. $$
I have the feeling that there is a simple way to show that the two expressions are the same with $L= \tilde L$. Can somebody point me in the right direction?

Denoting 
\begin{equation}
 h(x)=x^{l+1}e^{-x/2};\quad \varphi(x)=x^{2l+1}e^{-x}; \quad\Theta=x \frac{d}{dx}-x \frac{d^2}{dx^2}  + \frac{l(l+1)}{x} -\frac{x}{4}
\end{equation} 
both polynomial definitions can be written as 
\begin{align}
 \tilde{L}^{(2l+1)}_n(x) &= \frac{1}{n!h(x)}\Theta^n\left[h(x)\right]\\
 L^{(2l+1)}_n(x) &= \frac{1}{n!\varphi(x)}\frac{d^n}{x^n}\left[x^n\varphi(x)\right]
\end{align} We introduce the function $$F(x)=x^{-l}e^{x/2}$$ 
It can easily be checked that $\Theta\left[F(x)\right]=0$. Then, for any function $z(x)$,
\begin{equation}
 \Theta\left[F(x)z(x)\right]=xF(x)z'(x)-xF(x)z''(x)-2xF'(x)z'(x)
\end{equation} 
But $F'(x)=\frac{1}{2}F(x)-\frac{l}{x}F(x)$, hence
\begin{align}
 \Theta\left[F(x)z(x)\right]&=F(x)\left[2lz'(x)-xz''(x)\right]\\
 &=F(x)\frac{d}{dx}\left[ 2l+1  -x\frac{d}{dx}\right]z(x)
\end{align} 
Denoting
\begin{equation}
 M= 2l+1  -x\frac{d}{dx}
\end{equation} 
 we have shown that
\begin{equation}
 \Theta\left[F(x)z(x)\right]=F(x)\frac{d}{dx}M\left[z(x)\right]
\end{equation} 
and thus, for $n=1,2,3\ldots$,
\begin{equation}
 \Theta^n\left[F(x)z(x)\right]=F(x)\left( \frac{d}{dx}M \right)^n\left[z(x)\right]
\end{equation} 
Choosing $z(x)=\varphi(x)=x^{2l+1}e^{-x}$, we have $F(x)\varphi(x)=h(x)$ and then
\begin{equation}
 \Theta^n\left[h(x)\right]=F(x)\left( \frac{d}{dx}M \right)^n\left[\varphi(x)\right]
\end{equation} 
or, dividing both sides by $n!h(x)$,
\begin{equation}
 \frac{1}{n!h(x)}\Theta^n\left[h(x)\right]=\frac{1}{n!\varphi(x)}\left( \frac{d}{dx}M \right)^n\left[\varphi(x)\right]
\end{equation} 
It can be checked that $M\left[\varphi(x)\right]=x\varphi(x)$ which gives, for $n=1$,
\begin{equation}
 \frac{1}{h(x)}\Theta\left[h(x)\right]=\frac{1}{\varphi(x)}\frac{d}{dx}\left( x\varphi(x) \right)
\end{equation} 
and thus $\tilde{L}^{(2l+1)}_1(x)=L^{(2l+1)}_1(x)$. To go further, we use induction. Suppose that for some $n$
\begin{equation}
 \left( \frac{d}{dx}M \right)^n\left[\varphi(x)\right]=\frac{d^n}{dx^n}\left( x^n\varphi(x) \right)
\end{equation} 
then
\begin{equation}
 \left( \frac{d}{dx}M \right)^{n+1}\left[\varphi(x)\right]= \frac{d}{dx}M \frac{d^n}{dx^n}\left( x^n\varphi(x) \right)
\end{equation} 
For any function $z(x)$,
\begin{align}
 M\frac{d}{dx}z(x)&=\left( 2l+1 -x\frac{d}{dx}\right)\frac{d}{dx}z(x)\\
 &=\frac{d}{dx}\left(2l+2-x\frac{d}{dx}\right)z(x)\\
 &=\frac{d}{dx}\left( M+1 \right)z(x)
\end{align}
More generally, 
\begin{equation}
 M\frac{d^n}{dx^n}=\frac{d^n}{dx^n}\left( M+n \right)
\end{equation} 
Then
\begin{align}
 \left( \frac{d}{dx}M \right)^{n+1}\left[\varphi(x)\right]&=\frac{d^{n+1}}{dx^{n+1}}\left( M+n \right)\left( x^n\varphi(x) \right)\\
 &=\frac{d^{n+1}}{dx^{n+1}}\left( 2l+n+1-x\frac{d}{dx} \right)\left(x^{2l+n+1}e^{-x} \right)\\
 &=\frac{d^{n+1}}{dx^{n+1}}\left( x^{2l+n+2}e^{-x} \right)\\ 
 &=\frac{d^{n+1}}{dx^{n+1}}\left( x^{n+1}\varphi(x) \right)\\ 
\end{align}
As the induction hypothesis is true for $n=1$, we have shown that
\begin{equation}
 \frac{1}{n!h(x)}\Theta^n\left[h(x)\right]=\frac{1}{n!\varphi(x)} \frac{d^n}{dx^n}\left(x^n\varphi(x)\right)
\end{equation} 
and thus $\tilde{L}^{(2l+1)}_n(x)=L^{(2l+1)}_n(x)$, both series of polynomials are identical.
