Umbral calculus - eigenfunctions of operator

Begin with the study of polynomials in $X$ and some linear operators. Define the shift linear operator $$ E_h[f(X)] \!:=\! f(X\!+\!h), \tag{1} $$ and the difference linear operator $$ \Delta_h[f(X)] := f(X\!+\!h)\!-\!f(X), \tag{2} $$ and the derivative linear operator $$ D[f(X)] := \frac{d}{dX} f(X), \tag{3} $$ where $\,f()\,$ is any polynomial.

Define the falling factorial linear operator for monomials $$ L[X^n] := X(X-1)\dots(X-n+1). \tag{4} $$ Without loss of generality, define the $\,S\,$ linear operator by $$ S[f(X)] := (X\Delta_{-1}+c\,\Delta_{1})[f(X)]. \tag{5}$$ Applying this to falling factorials gives $$ S[(X)_n] = -n ( (X)_n - c\, (X)_{n-1}). \tag{6} $$ Rewriting this using the $\,L\,$ operator gives $$ S[L[X^n]] = -n L[ X^{n-1} (X-c)]. \tag{7} $$

Rewrite this using the derivative operator gives $$ S[L[f(X)]] = -L[ D[f(X)](X-c)]. \tag{8} $$ Apply this to the case $\,f(X)=E_h[X^n]\,$ to get $$ S[L[E_h[X^n]]] = -L[D[E_h[X^n]](X-c)]. \tag{9} $$ Apply this to the case $\,h=-c\,$ to get $$ S[L[(X-c)^n]] = -L[D[(X-c)^n](X-c)]. \tag{10} $$ But we know the derivative of $\,(X-c)^n\,$ and so get $$ S[L[(X-c)^n]] = -n\,L[(X-c)^n]. \tag{11} $$ Define the eigenvector function $$ \psi_n(X) \!:=\! L[(X\!-\!c)^n] \!=\! \sum_{m=0}^n {n \choose m} (-c)^m(X)_{n-m}. \tag{12} $$ with eigenvalue $\,-n\,$ and is expanded into a finite sum using the binomial theorem.