Product of polynomial coefficients of a recurrence A recurrence is given by 
$f[0]=2x$, $f[1]=3x^3-x^2+x+1$, 
$$ 
f[n]=(x^{2^n}+1)f[n-1]+(x^{2^n}+1)(x^{2^n-1}+1)
$$
How does the PRODUCT of the nonzero coefficients of $f[n]$ scale with $n$?
 A: It follows by induction from the recursion that for all $n\ge2$ the polynomial $f_n$ writes as $f_n=(x^{2^{n}}+1)(x^{2^{n-1}}+1)g_n $ where $g_n$ satisfies
$$\begin{cases} g_2:=3x+2 \\ g_n:=(x^{2^{n-2}}+1)g_{n-1}+1,\quad & \mathrm{if }\; n>2\end{cases}$$
and has degree $2^{n-1}-1$. Since $f_n=(x^{2^{n}}+1)(x^{2^{n-1}}+1)g_n $, the list of coefficients of $f_n$ is that of $g_n$ repeated $4$ times, and their product is the fourth power of the product $P_n$ of the coefficients of  $g_n$.  Also, the list of coefficients of $g_n$ is the one of $g_{n-1}$ repeated twice, with the constant term incremented by one. Thus the constant term is $g_n(0)=n$ and
$$\begin{cases} P_2:=3\cdot 2, \\ P_n:= {n\over n-1}\ P_{n-1}^2\quad & \mathrm{if }\; n>2\end{cases}$$
whence it follows by induction, for $n\ge2$
$$P_n=3^{2^{n-2}}n\prod_{k=1}^{n-1}k^{2^{n-1-k}}.$$
So the product of coefficients of $f_n$ is $P_n^4=3^{2^{n}}n^4\prod_{k=1}^{n-1}k^{2^{n+3-k}}$, and as to how it scales with $n$ I'd say it is quite a lot larger. For a more precise asymptotics, $$\log P^4_n=2^{n}\bigg({\log3}  +8\sum_{k=1}^{n-1}{\log k\over 2^k}+{4\log n\over 2^{n}}\bigg)=2^{n}\bigg({\log3}  +8\sum_{k=1}^{\infty}{\log k\over 2^k} +o(1)\bigg).$$
[edit] sorry for mis-reading  the correct definition of the iteration. The correct result is kindly given by მამუკა ჯიბლაძე in a comment below!
