I would guess with you that it isn't true, but would not be surprised if it is.

I wanted to ask how much you know and provide this, for what it is worth:

My calculations, **which I would need to repeat to be confident of**, are  that there are a few, but not that many, ways to pick the signs out past $\pm x^{N}$ for $N$ around $15$ so that the coefficients of the product are all $0,\pm1$ out to the $x^{N}$ term. For example one must start $1-x-x^2$ and the next thing must be $+x^{3}$ because otherwise the first term is $$1-x-x^2-x^3+x^4+x^5+x^6+x^7-x^8+x^9$$$$-x^{10}+x^{11}-x^{12}-x^{13}-x^{14}-x^{15}-x^{16}+cx^{17}-x^{18}$$  $$-cx^{19}-cx^{20}-cx^{21}+x^{22}+x^{23}+x^{24}+dx^{25}+\cdots$$ which leaves $$(c+d-4)x^{25}$$ in the product.

I did not as carefully check the continuations of $1-x-x^2+x^3$ although things were not quite as constrained.


Again, I am not confident that the above is error free, but the approach seems feasible. Is this in line with what you did and, if so, how far did you push it out?