If we consider the product: $$\prod_{i \ge 1}(1-a_ix^i)=P_n(x)$$ where $P_n(x)$ is polynomial with $n$ roots. It is obvious that the zeros of product are not zeros of polynomial because they are infinite while zero count of right side is $n$. My explanation is that the zeros is out of radius of convergence. Is this correct conclusion or there is something else?
ADDITION: In general if we have $$\prod_{i \ge 1}(1-a_ix^i)=f(x)$$ is all roots of product is also root of $f(x)$ ? Answer is no. In this case what is the reason of this "inconsistence"?
ADDITION: One example: $$\prod_{k >= 1}{(1-a_kx^k)} = 1 - \frac{x}{1-x}$$ $a_1=1$, $a_2=1$, $a_3=2$, $a_4=8$,...
I can think of some cases that work. Since $$ \prod_{j=0}^\infty \left(1 + z^{2^j}\right) = \sum_{n=0}^\infty z^n = \frac{1}{1-z}$$ if $N$ is not a power of $2$ we have $$ P(z) = \sum_{n=0}^{N-1} z^n = \frac{1-z^N}{1-z} = \prod_{k=1}^\infty (1 + a_k z^k)$$ where $$\eqalign{a_{2^j} &= 1\cr a_N &= -1\cr a_k &= 0\ \text{otherwise}}$$ all converging in the open unit disk. Indeed, the roots of $1 + z^{2^j}$ are not roots of $P(z)$.