Suppose function $f$ is such that for any $x\in R_+$ we have that $$ f(\, x(x-1)\ldots(x-k+1)\, )=A_k. $$ Moreover, the following holds for all $k$: $$ f(x)=A_1\\ f(x(x-1))=f(x^2-x)=f(x^2)-f(x)=A_2\\ f(x(x-1)(x-2))=f(x^3-3x^2+2x)=x^3-3f(x^2)+2f(x)=A_3\\ \ldots $$ Question: I would like to find a closed formula, or at least an upper bound of $f(x^n)$ in terms of $A_i$.
My attempt: I have started representing $f(x^n)$ as following, but I cannot find a pattern: $$ f(x)=A_1\\ f(x^2)=A_2+f(x)=A_2+A_1\\ f(x^3)=A_3-3f(x^2)-2f(x)=A_3+3A_2+3A_1-2A_1=A_3+3A_2+A_1\\ \ldots $$
Because that finite product is $$x(x-1)\cdots (x-k+1)=\sum_{m=0}^k S_k^m x^m,$$ which is basically a list of Stirling numbers of the first kind, you are essentially trying to invert that matrix of the Stirling numbers of the first kind and get the Stirling numbers of the second kind, with coefficients listed e.g. in https://oeis.org/A008277.