I have conjectured that for all functions $$ g(t) = \sum_{n=0}^{\infty}\frac{a_n}{n!}t^n, $$ it holds that $$ \int_0^{\infty}\frac{(g(tx)-g(0))^j}{j!}e^{-t} dt = \frac{1}{j!a_0a_1...a_j}\prod_{r=1}^j(\int_0^{\infty}g(rt_rx)e^{-t_r}dt_r - g(0)), $$ provided that the product of coefficients only includes the non-negative coefficients. This is a generalization of the generating function for Stirling numbers of the second kind: $$\sum_{n=j}^{\infty} {n \brace j}x^n = \prod_{r=1}^j(\frac{1}{1-rx}-1)\frac{1}{r}, $$ which is acheived when $g=exp$.
I was able to arrive at this by inductive reasoning, i.,e., I looked at the Stirling sum and tried experimenting with it. However, the equality seems to hold up to a constant. Both functions are correct, except the coefficient of the leading term of the product is different. This can be seen in the case of cosine, for example, or the square of cosine.
Is there a way to correct this leading coefficient such that this equality would hold without any issues?