Suppose that $f,g\_1,g\_2,\dots$ are functions from $\mathbb{R}$ to $\mathbb{R}$ such that $f'=f\,g\_1$ and $g'\_j=g\_j^2-g\_j g\_{j+1}$. Here and in what follows, $j$ is any natural number. Then, by induction, $f^{(j)}=f\, P\_j(g\_1,\dots,g\_j)$, where 
$$P\_j(u\_1,\dots,u\_j)=\sum\_{k\_1=0}^j\cdots\sum\_{k\_j=0}^j c\_{k\_1,\dots,k\_j}
u\_1^{k\_1}\cdots u\_j^{k\_j}
$$
is a polynomial in $\mathbb{R}[u\_1,\dots,u\_j]$ of degree $j$. 


The problem is to show that 
$$\sum\_{k\_1=0}^j\cdots\sum\_{k\_j=0}^j |c\_{k\_1,\dots,k\_j}|\, 
1^{k\_1}2^{k\_2}\cdots j^{k\_j}=2^{j - 1} j! 
$$
for all natural $j$. 
This equality has been verified for $j=1,\dots,10$.



A property of the derivatives of a function
We have, writing for short $P\_n=P\_n(g\_1,\dots,g\_{n })$,


$$\big(f^{(n)}\big)' =\big(fP\_{n }\big)'=f' P\_{n }+\sum\_{i=1}^n f\partial\_iP\_ng\_i'=fg\_1 P\_{n }+\sum\_{i=1}^n f\partial\_iP\_n(g\_i^2-g\_ig\_{i+1})$$
Comparing with $f^{(n+1)}=fP\_{n+1}$ we obtain a linear recursion for the sequence of polynomials $P\_n=P\_n(x\_1,\dots,x\_{n })$
\begin{cases} P\_1=x\_1 \\ P\_{n+1}=x\_1P\_n +\sum\_{i=1}^n (x\_i - x\_{i+1}) x\_i\partial\_i P\_n, \end{cases} 
whence it is clear that $P\_n$ is a homogeneous polynomial of degree $n$, $P\_ n:=\sum\_{|\alpha|=n} c(\alpha)x^\alpha$, the sum being extended over all multi-indices $\alpha:=(\alpha\_1,\alpha\_2,\dots )$ of weight 
$|\alpha|:=\alpha\_1+\alpha\_2+\dots=n.$


Equating the coefficients of the monomial $x^\alpha:=x\_1^{\alpha\_1}\dots x\_{n+1}^{\alpha\_{n+1}} $ of degree $n+1:=|\alpha|$ in the above recursive formulas we obtain:
$$c(\alpha)= \alpha\_1 c(\alpha-\delta\_1)- \sum\_{i=2}^{n} (\alpha\_{i-1}-\alpha\_{i }+1)c(\alpha-\delta\_i) \; - \alpha\_{n}c(\alpha-\delta\_{n+1}),$$
where $\delta\_i:=(\delta\_{i1},\delta\_{i2},\delta\_{i3}\dots)$ and $\delta\_{ij}$ is the usual Kronecker symbol.
From this formula for $c(\alpha)$ it follows easily by induction that


**(i)** If $c(\alpha)\neq0$ then $\alpha\_1\ge\alpha\_2\ge \dots $;


**(ii)** $\operatorname{sgn}c(\alpha)=(-1)^{|\alpha|-\alpha\_1}$.


As a consequence, the sum we are interested in is $$ \sum\_{|\alpha|=n} \big|c(\alpha)\big|1^{\alpha\_1}2^{\alpha\_2}\dots n^{\alpha\_n}=P\_n( 1, - 2,- 3,\dots,- n).$$


To evaluate it, consider $\sigma\_n(t):=P\_n( t, - 2,- 3,\dots,- n).$ Then, by the recurrence relation of the polynomials $P\_n$ 
and by the Euler formula for homogeneous polynomials we find
\begin{cases} \sigma\_1=t \\ \sigma\_{n+1} =(t+n)\sigma\_n + (t+t^2)\sigma\_n' , \end{cases} 


Now it is easy to check that
$$\sigma\_n(t)=n!t(t+1)^{n-1},$$
whence $\sigma\_n(1)=n!2^{n-1}.$ 


$$\*$$


**rmk.** Also note that the polynomials $Q\_n:= \sum\_{|\alpha|=n} \big|c(\alpha)\big|x^\alpha$ are simply $P\_n(x\_1, -x\_2,-x\_3,\dots,-x\_n),$ whence one finds a linear recurrence relation for them 


\begin{cases} Q\_1=x\_1 \\ Q\_{n+1}=x\_1Q\_n+2x\_1^2\partial\_1Q\_n +\sum\_{i=1}^n (x\_{i+1} - x\_{i}) x\_i\partial\_i Q\_n. \end{cases} 


