Evaluating the recurrence $f_k(x)= f'_{k-1}(x)+f_{k-1}(x) f_1(x)$ with $f_0(x)=1$, $f_1(x)=e^x$

It looks like $$ f_n(x) = \sum_{k=0}^{n} \left\{{n \atop k}\right\} e^{kx}, $$where $\left\{{n\atop k}\right\}$ are the Stirling numbers of the second kind. You can show these satisfy the usual recurrence using basic differentiation rules.

As far as generalizing, i.e. determining $f_{1/2}(x)$, this can be done several ways, for instance in this paper. Hope that helps!

More explicitly,

$f_{k-1}'(x) =f_k(x)-f_{k-1}(x) f_1(x) =f_k(x)-f_{k-1}(x)e^x $ so $f_k(x) =f_{k-1}'(x)+f_{k-1}(x)e^x $ or, shifting the index and removing $(x)$ for conciseness, $f_{k+1} =f_{k}'+e^xf_{k} $.

If $f_k(x) =\sum_{j=1}^k c_{k, j}e^{jx} $, with $c(1, 1) = 1$, then

$f_k'(x) =\sum_{j=1}^k jc_{k, j}e^{jx} $

so

$\begin{array}\\ f_{k+1} &=f_{k}'+e^xf_{k}\\ &=\sum_{j=1}^k c_{k, j}(e^{jx})'+e^x\sum_{j=1}^k c_{k, j}e^{jx}\\ &=\sum_{j=1}^k jc_{k, j}e^{jx}+\sum_{j=1}^k c_{k, j}e^{(j+1)x}\\ &=\sum_{j=1}^k jc_{k, j}e^{jx}+\sum_{j=2}^{k+1} c_{k, j-1}e^{jx}\\ &=c(k, 1)e^x+\sum_{j=1}^k (jc_{k, j}+c(k, j-1))e^{jx}+c_{k, k}e^{(k+1)x}\\ &=\sum_{j=1}^{k+1} c_{k+1, j}e^{jx}\\ \end{array} $

so

$c(k+1, 1) = c(k, 1) = 1 $, $c(k+1, k+1) = c(k, k) =1 $, and, for $j = 2$ to $k$, $c(k+1, j) =(jc_{k, j}+c(k, j-1)) $

and these turn out to be the Stirling numbers of the second kind as Integrand wrote.