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I attempt to understand the $0$-dimensional QFT from these QFT lecture notes by Ronald Kleiss from 2019. The author defines the generating function $Z(J)$ and its logarithm in the following way.

$$Z(J) = \sum_{n\geq 0} G_n \frac{J^n}{n!}, \quad W(J) = \ln(Z(J)) = \sum_{n\geq 0} C_n \frac{J^n}{n!}.$$

Exercise 1 (page no. 59) is to solve $C_n$s in terms of $G_n$s. For it, I proceed in the following way. \begin{eqnarray} W(J) = \ln(Z(J)) &=& \ln\left(G_0 - \sum_{n = 1}^{\infty} (-G_n) \frac{J^n}{n!}\right) \\ &=& \ln\left(1 - \underbrace{\sum_{n = 1}^{\infty} (-G_n) \frac{J^n}{n!}}_{\equiv x}\right) \\ \\ &=& \sum_{s=1}^{\infty} \frac{1}{s} \left( \sum_{n = 1}^{\infty} G_n \frac{J^n}{n!}\right)^s = \sum_{n=0}^{\infty} C_n \frac{J^n}{n!}, \tag{1} \end{eqnarray} where I have used the given fact that $G_0 = 1$ and $\ln(1-x) = \sum_{s=1}^{\infty} \frac{-x^s}{s}$.

From Eq. (1) I deduce $$C_s = \frac{s!}{J^s s} \left[ \sum_{n = 1}^{\infty} G_n \frac{J^n}{n!}\right]^s.$$

This formula for high values of $s$; e.g., $s =2, 3, 4$, etc. is not very helpful. How can I solve the problem in a more efficient way?

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An efficient approach is to realise the $G_n$ are cycle index polynomials of the symmetric group with arguments $C_n$. This is more efficient because properties of these polynomials are well studied: see, eg, Appendix B in:

https://arxiv.org/abs/1912.01055

for their definition (eqn B.675) and derivations of many of their properties. So the general relation is: $$ \frac{1}{n!}G_n(C_s)=e^{C_0}Z_n[C_s/(s-1)!] $$

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From the given expression for $C_n$ we can write, $$C_n = \left[ \frac{\partial^n W(J)}{\partial J^n}\right]_{J=0},$$ where $W(J) = \ln Z(J) = \ln \left[\sum_{n=0}^{\infty}\frac{1}{n!} J^n G_n\right]$.

Then determine different $C_n$ s.

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