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?

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

$$ L^k_n(x) = frac{e^x}{n!x^k} D^n[x^{n+k}/e^x]. tag{10} $$