What is the solution to $n_{i}=(e^{-x})\frac{dn_{i-1}}{dx}$ for i=0,1,2,...? Given the recursion relationship $n_{i+1}=(e^{-x})\frac{dn_{i}}{dx}$ for $i=0,1,2,\ldots,$ where $n_{0}$ is a known function of $x$, what is the solution to $n_{k}$ in terms of $x$ and the derivatives of $n_{0}$, where $k$ is any positive integer?
Re what I've tried, things simplify a bit if I express $n_0$ as $n_0=e^{y}$ and look for the solution in terms of the derivatives of $y$ (which may be expressed in terms of $x$, $n_0$, and the derivatives of $n_0$ as required by the question). We have then,
$n_1=e^{y-x}\frac{dy}{dx}$
$n_{2}=e^{y-2x}\left(  \frac{d^{2}y}{dx^{2}}+\left(  \frac{dy}{dx}\right)^{2}-\frac{dy}{dx}\right)$
$n_{3}=e^{y-3x}\left(  \frac{d^{3}y}{dx^{3}}+3\frac{dy}{dx}\frac{d^{2}%
y}{dx^{2}}+\left(  \frac{dy}{dx}\right)  ^{3}-3\left(  \frac{dy}{dx}\right)
^{2}-3\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}\right)  $
This gives a feel for where things are headed. The solution comes down to figuring out the integer coefficients of all the resultant derivatives and products of derivatives.
 A: I will solve the slightly more general recurrence problem
$$n_k=e^{\lambda x}\frac{d}{dx}n_{k-1}$$
Denote $f(x):=e^{\lambda x}, D:=d/dx$ for brevity. Clearly,  $Df=\lambda f$ for exponentials are eigenfunctions of the derivative operator. By induction, it can be easily shown that the solution to the recurrence takes the form
$$n_k(x)=f^k(x)\lambda^k~ \left[\Omega_k \left(\frac{D}{\lambda}\right)\right]n_0(x)$$
where $\Omega_k$ are polynomials of degree $k$. The quantity in brackets is a finite sum of powers of the derivative operator, and hence well defined. To generate a recurrence for these polynomials, act with $fD$ on both sides of the equation and reuse it for the $k+1$-th term of the recurrence to obtain the very simple formula
$$\Omega_{k+1}(x)=(x+k)\Omega_k(x)$$
which is solved by the family of polynomials given by the rising factorial:
$$\Omega_k(x)=x^{(k)}=\prod_{m=1}^{k-1}(x+m)=\sum_{m=0}^k(-1)^{k-m}s(k,m)x^m$$
where $s(k,m)$ denote the Stirling numbers of the first kind. The coefficients in your expansion are proportional to them. More precisely, for $\lambda=-1$, it is seen that
$$n_k(x)=e^{-kx}\sum_{m=0}^k s(k,m)\frac{d^m}{dx^m}n_0(x)$$
and this concludes the analysis.
