Solving $left(x-c_1frac{d}{dx}right)^nf(x)=0$ for $f(x)$

Here is a fast track. Writing $D=partial/partial x$, $c=c_1$ we have the commutation relation: $$ ( x - c D) ; e^{frac{x^2}{2c}} = e^{frac{x^2}{2c}} (-cD) $$ Thus, $$ 0 = ( x - c D)^n f_n(x) = ( x - c D)^n e^{frac{x^2}{2c}} e^{-frac{x^2}{2c}} f_n(x) = e^{frac{x^2}{2c}} (- c D)^n e^{-frac{x^2}{2c}} f_n(x) Leftrightarrow $$ $$ (- c D)^n e^{-frac{x^2}{2c}} f_n(x) = 0Leftrightarrow$$ $$ e^{-frac{x^2}{2c}} f_n(x) = P_n(x), P_nin {Bbb R}_n[x]Leftrightarrow$$ $$ f = P_n(x)e^{frac{x^2}{2c}}, P_nin {Bbb R}_n[x].$$

A fairly general formula (less well-known but with the same proof) is obtained by considering $qin C^infty({Bbb R})$. Then $$ (D - q'(x))^n f_n(x)=0 Leftrightarrow f_n = P_n(x) e^{q(x)}, P_nin {Bbb R}_n[x].$$

Alternatively, let $D$ denote the differential operator $$Df:=f'$$ for all differentiable function $f:mathbb{R}tomathbb{C}$. Define the operator $M$ as $$(Mf)(x):=expleft(+dfrac{x^2}{2c_1}right),f(x)$$ for all $f:mathbb{R}tomathbb{C}$ and $xinmathbb{R}$. Observe that $M$ is an invertible operator with the inverse $M^{-1}$ given by $$(M^{-1}f)(x)=expleft(-dfrac{x^2}{2c_1}right),f(x)$$ for all $f:mathbb{R}tomathbb{C}$ and $xinmathbb{R}$. Now, we conjugate the differential operator $D$ by $M$ to obtain the operator $Delta:=MDM^{-1}$ which satisfies $$(Delta f)(x)=left(frac{text{d}}{text{d}x}-frac{x}{c_1}right),f(x)$$ for all differentiable function $f:mathbb{R}tomathbb{C}$ and $cinmathbb{R}$. Therefore, the question asks for all $n$-time differentiable functions $f:mathbb{R}tomathbb{C}$ in the kernel of $Delta^n$, namely, $$left(Delta^n fright)(x)=left(frac{text{d}}{text{d}x}-frac{x}{c_1}right)^n,f(x)=0$$ for all $xinmathbb{R}$. Now, observe that $$Delta^n=(MDM^{-1})^n=MD^nM^{-1},.$$ Thus, $fin ker(Delta^n)$ if and only if $M^{-1}finker(D^n)$. Since $ker(D^n)$ contains all polynomials of degree less than $n$, we conclude that there exists a polynomial function $p:mathbb{R}tomathbb{C}$ of degree less than $n$ such that $$expleft(-frac{x^2}{2c_1}right),f(x)=big(M^{-1}fbig)(x)=p(x),,$$ for each $xinmathbb{R}$. Thus, $$f(x)=(Mp)(x)=expleft(+frac{x^2}{2c_1}right),p(x)$$ for all $xinmathbb{R}$.

Furthermore, for any function $g:mathbb{R}tomathbb{C}$ with an $n$-th antiderivative, all solutions $f:mathbb{R}tomathbb{C}$ which are $n$-time differentiable and satisfy $$Delta^n f=g,,$$ or equivalently, $$left(frac{text{d}}{text{d}x}-frac{x}{c_1}right)^n,f(x)=g(x)$$ for all $xinmathbb{R}$, are given by $$f(x)=expleft(+frac{x^2}{2c_1}right),big(G(x)+p(x)big),,$$ for all $xinmathbb{R}$, where $G$ is an $n$-th antiderivative of $M^{-1}g$, and $p:mathbb{R}tomathbb{C}$ is a polynomial function of degree less than $n$. For example, one can take $$G(x):=int_0^x,int_0^{x_1},cdots,int_0^{x_{n-1}},int_0^{x_n},expleft(-frac{x_n^2}{2c_1}right),g(x_{n}),text{d}x_{n},text{d}x_{n-1},cdots, text{d}x_2,text{d}x_1,.$$

In general, if $h:mathbb{R}tomathbb{C}$ has a first antiderivative $H$, then all $n$-time differentiable functions $f:mathbb{R}tomathbb{C}$ such that $$left(frac{text{d}}{text{d}x}-h(x)right)^n,f(x)=0$$ for each $xinmathbb{R}$ take the form $$f(x)=expbig(+H(x)big),p(x)$$ for all $xinmathbb{R}$, where $p:mathbb{R}tomathbb{C}$ is a polynomial function of degree less than $n$. If $g:mathbb{R}tomathbb{C}$ has an $n$-th antiderivative, then all all $n$-time differentiable functions $f:mathbb{R}tomathbb{C}$ such that $$left(frac{text{d}}{text{d}x}-h(x)right)^n,f(x)=g(x)$$ for every $xinmathbb{R}$ take the form $$f(x)=expbig(+H(x)big),big(G(x)+p(x)big)$$ for all $xinmathbb{R}$, where $p:mathbb{R}tomathbb{C}$ is a polynomial function of degree less than $n$ and $G(x)$ is the $n$-th antiderivative of $expbig(-H(x)big),g(x)$. We may take $$G(x):=int_0^x,int_0^{x_1},cdots,int_0^{x_{n-1}},int_0^{x_n},expbig(-H(x_n)big),g(x_{n}),text{d}x_{n},text{d}x_{n-1},cdots, text{d}x_2,text{d}x_1,.$$

One can solve it recursively. For example, let $f_n(x)$ be such that $$left(x-c_1frac{d}{dx}right)^nf_n(x)=0.$$ Then one needs to find $f_{n+1}(x)$ such that $$left(x-c_1frac{d}{dx}right)^{n+1}f_{n+1}(x)=0$$ $$Leftrightarrow left(x-c_1frac{d}{dx}right)^nleft[left(x-c_1frac{d}{dx}right)f_{n+1}(x)right]=0$$ $$Leftrightarrow left(x-c_1frac{d}{dx}right)f_{n+1}(x)=f_n(x),$$ where the latter can be solved by the standard method (e.g. using integrating factor) to yield $$f_{n+1}(x)=e^{frac {x^2}{2c_1}}intfrac{f_n(x)}{-c_1}e^{-frac{x^2}{2c_1}}~dx.quad (1)$$

As Maxim observed in the comment, the answer turns out to be simple. No Hermitian polynomials are needed. Let $D=x-c_1frac{d}{dx}$. There case $n=0$ being trivial, one needs to check that the solutions to $$D^nf(x)=0,ngeq 1$$ are given by $$f_n(x)=p(x)e^{frac {x^2}{2c_1}},$$ where $p(x)$ is a polynomial with $deg pleq n-1.$

This can be proved by induction. You already obtained the case $n=1$. Assume the result is true for some $ngeq 1$, so the solutions to $D^nf(x)=0$ is of the form $$f_n(x)=p(x)e^{frac{x^2}{2c_1}},deg pleq n-1.$$ Now by (1), one solves the equation $D^{n+1}f=0$ and obtains (up to constant multiple) $$f_{n+1}(x)=e^{frac{x^2}{2c_1}}int p(x)~dx=q(x)e^{frac{x^2}{2c_1}},$$ for some polynomial $q(x)$ with $deg qleq n.$ QED