How is the Rodrigues formula $L_n^k(x)=\frac{e^x x^{-k}}{n!}\frac{d^n}{dx^n}(e^{-x}x^{n+k})$ derived? I am trying to deduce the Rodrigues formula for generalized Laguerre polynomials $$L_n^k(x)=\frac{e^x x^{-k}}{n!}\frac{d^n}{dx^n}(e^{-x}x^{n+k})$$ but I have reached a point where I do not know how to proceed, my procedure was as follows:
I started from the following two equalities:

*

*$L_n^k(x)=(-1)^{k}\cfrac{d^k}{dx^k}L_{n+k}(x)$

*$L_n(x)=\cfrac{e^x}{n!}\cfrac{d^n}{dx^n}(x^ne^{-x})$
Then substituting $ n + k $ for $ n $ in equation (2) we obtain that:
$$L_{n+k}(x)=\frac{e^x}{(n+k)!}\frac{d^{n+k}}{dx^{n+k}}(x^{n+k}e^{-x})$$
And therefore, substituting the value of $ L_ {n + k} $ in equation (1) we find that:
$$L_n^k(x)=(-1)^{k}\frac{d^k}{dx^k}\cfrac{e^x}{(n+k)!}\frac{d^{n+k}}{dx^{n+k}}(x^{n+k}e^{-x})$$
But I no longer know how to continue.
I would really appreciate your help.
 A: Given:
$$ L^k_n(x) := (-1)^k D^k[ L_{n+k}(x)] \tag{1} $$
using the linear differential operator $\,D[f(x)]:=f'(x)\,$ and where
$$ L_n(x) := e^x/n! D^n[ x^n/e^x ]. \tag{2} $$
Rewrite equation $(2)$ as
$$ L_n(x) = T^n[ x^n ]/n! \tag{3} $$
where the linear differential operator $\,T\,$ is defined as
$$ T[f(x)] := f'(x)-f(x) = (D-I)[f(x)] \tag{4} $$
and because of the identity
$$ T[ f(x) ] / e^x = D[ f(x)/e^x ]. \tag{5} $$
Combine equations $(3)$ and $(1)$ to get
$$ L^k_n(x) = (-1)^k D^k[ T^{n+k}[ x^{n+k} ] / (n+k)! ]. \tag{6} $$
The differential operators $\,D\,$ and $\,T\,$ commute, implying
$$ L^k_n(x) = (-1)^k T^{n+k}[ D^k[ x^{n+k} ] / (n+k)! ]
  = (-1)^k T^{n+k}[ x^n/n! ]. \tag{7} $$
Due to the binomial expansions of $\,T^n\,$ and $\,T^{n+k}\,$ we get
$$ T^n[ x^{n+k} ] = (-1)^k T^{n+k}[ x^n] x^k \tag{8} $$
by expanding both sides into polynomials and
comparing their coefficients.
Substitute this equation $(8)$ into equation
$(7)$ to get
$$ L^k_n(x) = T^n[ x^{n+k}]/(n!x^k). \tag{9} $$
Rewrite this equation $(9)$ using equation $(5)$
to get
$$ L^k_n(x) = \frac{e^x}{n!x^k} D^n[x^{n+k}/e^x]. \tag{10} $$
