Simplify recurrence $\frac{d}{dx} f_{n-1}(x)= f_n(x)- f_{n-1}(x) f_1(x)$ Suppose we have a sequence of infinitely differentiable functions $ \{ f_k(x) \}$.   Now suppose that these functions satisfy the following recursion:
\begin{align}
\frac{d}{dx} f_{n-1}(x)= f_n(x)-  f_{n-1}(x) f_1(x)
\end{align}
where $f_0(x)=1$ for all $x$.  Can we re-write $f_n(x)$ only in terms of $f_1(x)$ and derivatives of $f_1(x)$?
It is not difficult to see that this is possible. The difficulty is to create the exact formula.
 A: $$\frac{d}{dx}f_{n-1}(x)=f_n(x)-f_{n-1}(x)f_1(x),\ \ \ \ f_0(x)=1$$
$$f_n(x)=\frac{d}{dx}f_{n-1}(x)+f_{n-1}(x)f_1(x)$$
$\ $
$$f_1(x)=\frac{d}{dx}f_0(x)+f_0(x)f_1(x)=f_1(x)$$
$$f_2(x)=\frac{d}{dx}f_1(x)+f_1(x)f_1(x)$$
$$f_3(x)=\frac{d}{dx}f_2(x)+f_2(x)f_1(x)$$
$$...$$
$\ $
$f_n(x)$ is the following special complete exponential Bell polynomial:
$$f_n(x)=B_n(f_1^{(0)}(x),f_1^{(1)}(x),...,f_1^{(n-1)}(x))=\sum_{\sum_{t=1}^{n}tk_t=n}\frac{n!}{\prod_{i=1}^{n}i!^{k_{i}}k_{i}!}\prod_{i=1}^{n}{f_1^{(i-1)}(x)}^{k_{i}}$$
The vectors of running indices $(k_1,...,k_n)$ in the sum expression represent the integer partitions of $n$.
Applying Faà di Bruno's Formula (Higher chain rule), we get:
$$f_n(x)=e^{-\int f_1(x)dx}\frac{d^n}{dx^n}e^{\int f_1(x)dx}=e^{-\int f_1(x)dx}\frac{d^{n-1}}{dx^{n-1}}\left(f_1(x)e^{\int f_1(x)dx}\right)$$
A: I don't know if this is what you want. It is easy to see
$$f_n(x) = f_{n-1}^\prime(x) + f_{n-1}(x)f_1(x). \tag{1}$$
Multiplying both sides of (1) by $e^{\int f_1(x)dx}$ gives
$$f_n(x)e^{\int f_1(x)dx} = \bigg[f_{n-1}(x) e^{\int f_1(x)dx}\bigg]'. \tag{2}$$
Let
$$ g_n(x)=f_n(x)e^{\int f_1(x)dx}. $$
Then (2) gives
$$ g_n(x)=g_{n-1}'(x) $$
Using this recursion $(n-1)$-times, one has
$$ g_n(x)=g_1^{(n-1)}(x)=\frac{d^{n-1}}{dx^{n-1}}\bigg[f_1(x)e^{\int f_1(x)dx}\bigg] $$
and hence
$$ f_n(x)=e^{-\int f_1(x)dx}\frac{d^{n-1}}{dx^{n-1}}\bigg[f_1(x)e^{\int f_1(x)dx}\bigg] $$
