Fix $n>1$, I want to find power series $f(x) \in \mathbb{Q}[[x]]$ which satisfies $$ f((1+x)^n -1)= nf(x).$$
I know that the $0$ function and $\ln(1+x)$ satisfy this equation for all $n$.
My question is, if we fix $n$, does there exist more functions which are solutions? If so, can we characterise these functions depending on $n$? If not, can one give a proof that the above solutions are the only solutions for any $n>1$.
For any $n$, if you write $f(x) = a_1 x + a_2 x^2 + \ldots$ and substitute in to the difference of the two sides of the equation, the coefficients of each power $x^k$ are linear combinations of $a_1, a_2, \ldots, a_k$ that must be $0$: the coefficient of $x^1$ is $0$, of $x_2$ is $n (n-1) (a_1 + 2 a_2)/2$, etc. The coefficient of $x^m$ will be a combination of $a_1$ to $a_m$, with the coefficient of $a_m$ in this coefficient being $n^m-n$. Thus for any $n > 1$, it is possible to solve for $a_2, a_3, \ldots, $ in terms of $a_1$. We know the solution since $a_1 \ln(1+x)$ satisfies the equation. Thus all the solutions are $c \log(1+x)$ for arbitrary constant $c$.