Convergence issues with infinite product of formal series

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.

Just define the RHS as $\sum c_ix^i$, where $$ c_i = \sum_{\substack{i_1 < \dots < i_k \\ s_{i_1} + \dots s_{i_k} = i}} (-1)^k. $$ Then $c_i$ is a finite sum. It's clear that $|c_i| \le p(i) \ll (1+\epsilon)^i$ for all $\epsilon > 0$, where $p(i)$ is the partition function. Therefore, the RHS as defined converges absolutely for all $|x| < 1.$ This allows us to substitute values $x = 1 - \delta_i$ for $\delta_i \rightarrow 0$, and compute the RHS as written. As $x \rightarrow 1$, the RHS $\rightarrow 0$, so $P(x) \rightarrow 0$, so $P(1) = 0.$ Let $P(x) = (1-x)Q(x).$ Considering only values $|x| < 1$ still, we can multiply both sides by $(1+x+\dots)$ formally, and continue taking limits as $x \rightarrow 1.$ Repeating this argument, we can finish.