TITLE: Convergence issues with infinite product of formal series
QUESTION [6 upvotes]: Question first:

Show that if $s_1 < s_2 < \dots$ is an increasing sequence of positive integers and $P(x)$ is a nonzero polynomial then we cannot have
  $$ P(x) \equiv \prod_{j=1}^\infty (1 - x^{s_j}) $$
  as formal series.

The right-hand side really means $\lim_{N \to \infty} \prod_{j=1}^N (1 - x^{s_j})$ where the notion of convergence is as described in this math.SE answer by Bill.
The intuition is that if we take $x \to 1^-$ then the left-hand side tends to zero at a slower rate than the right-hand side, since $P$ has the root $1$ with only finite multiplicity, while every factor on the RHS tends to zero. However, since formal and functional power series aren't actually the same thing (see link above), I'm not sure how to make this precise, even after formally inverting the $1-x$ factors on the left-hand side (by multiplying both sides by $(1+x+x^2+\dots+)^k$).
Does anyone know the correct incantations?
(For context, this came up in a solution to a recent IMO proposal. Since it was for high school students I swept the convergence issues under the rug, but myself I'd like to know exactly what the right thing to do is.)

REPLY [4 votes]: Equivalently, we'll show that we cannot have
$$\frac{1}{P(x)} = \frac{1}{\prod_{j=1}^{\infty} (1 - x^{s_j})}$$
as formal power series. The idea is that the LHS has a pole of finite order at $x = 1$ while the RHS has an essential singularity at $x = 1$. Precisely, the coefficients on the LHS have asymptotic growth a polynomial times an exponential. On the other hand, the coefficients of the RHS can be shown to have growth both strictly larger than any polynomial (by truncating the product) and strictly smaller than any exponential (by comparing to the growth rate in the case where $s_j = j$, which is known). So the two rates of growth can't match.