Consider a more general problem on representing the formal power series $F:=c_0+c_1x+c_2x^2+\dots$ as the sum $$F=\sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n+1}(1-f(k)x).$$ We clearly have $a(0)=c_0$, and dividing by $1 - f(1)x$ and subtructing $a(0)$ we get the same kind problem of representing $G(x):=x^{-1}(F/(1-f(1)x)-c_0)$ as $$\sum\limits_{n=0}^{\infty}a(n+1)x^n\prod\limits_{k=1}^{n+1}(1-f(k+1)x).$$ The coefficients of the series $ G(x)=d_0+d_1x+d_2x^2+\dots $ satisfy the equations $$(d_0+d_1x+d_2x^2+\dots)(1-f(1)x)=(c_1+c_0f(1))+c_2x+c_3x^2+\dots ,$$ that is, $d_i-f(1)d_{i-1}=c_{i+1}$ for $i=0,1,\dots$, if we put $d_{-1}=c_0$. So, $(d)$ is obtained from $(c)$ by the first step of your algorithm. Then we proceed the same way.

Consider a more general problem on representing the formal power series $F:=c_0+c_1x+c_2x^2+\dots$ as the sum $$F=\sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n+1}(1-f(k)x).$$ (In your example all $c_i$ are equal to 1). We clearly have $a(0)=c_0$, and dividing by $1 - f(1)x$ and subtructing $a(0)$ we get the same kind problem of representing $G(x):=x^{-1}(F/(1-f(1)x)-c_0)$ as $$\sum\limits_{n=0}^{\infty}a(n+1)x^n\prod\limits_{k=1}^{n+1}(1-f(k+1)x).$$ The coefficients of the series $ G(x)=d_0+d_1x+d_2x^2+\dots $ satisfy the equations $$(d_0+d_1x+d_2x^2+\dots)(1-f(1)x)=(c_1+c_0f(1))+c_2x+c_3x^2+\dots ,$$ that is, $d_i-f(1)d_{i-1}=c_{i+1}$ for $i=0,1,\dots$, if we put $d_{-1}=c_0$. So, $(d)$ is obtained from $(c)$ by the first step of your algorithm for $\nu$. Then we proceed the same way.