Expansion of $(S_{I}[\bar{\alpha},\alpha])^n$ in the proof of the linked-cluster theorem

Question

On P.192 of Piers Coleman's "Introduction to many body physics", he uses Sam Edwards' replica trick to show that Green's function $$G(1-2)=\sum{\text{two-leg linked-cluster diagram}}\tag{7.72}$$ He first notes the equality $$\ln S_I=\lim_{n\rightarrow0}\bigg[\frac{{S_I}^n-1}{n}\bigg]\tag{7.69}$$ where $$S_I[\bar{\alpha},\alpha]=\langle\phi|T\exp(-i\int_{-\infty}^{\infty}dt(V(\psi^\dagger,\psi)+\text{source term}))|\phi\rangle$$ and $\alpha,\bar{\alpha}$ are source fields. Then note $$S_I(\bar{\alpha},\alpha)^n=\langle\phi|T\exp(-i\int_{-\infty}^{\infty}dt\sum^n_{\lambda=1}(V(\psi_\lambda^\dagger,\psi_\lambda)+\text{source term}))|\phi\rangle\tag{7.70}$$ where we've introduced $n$ identical but independent replicas of the original system, hence the label $\lambda$ on $\psi$. Finally, he states $$S_I^n=1+n[\text{linked-cluster diagrams}]+n^2[\text{disconnected diagrams that are products of two linked clustered diagrams}]+...$$ which is not obvious to me at all. This finishes his proof. My question is how to prove this expansion. I'm familiar with the proof of Feynman diagram that uses $$\int_tV(\psi^\dagger,\psi)\rightarrow \int_tV(\frac{\zeta\delta}{\delta\alpha(t)},\frac{\delta}{\delta\bar{\alpha}(t)})$$ to differentiate $\int d1d2\bar{\alpha}(1)G(1,2)\alpha(1)$. I assume we are now to expand $$\exp(\int_t\sum_{i}V(\frac{\zeta\delta}{\delta\alpha_i(t)},\frac{\delta}{\delta\bar{\alpha}_i(t)}))\exp(\sum_j\int d1_jd2_j\bar{\alpha}_j(1_j)G(1_j,2_j)\alpha_j(2_j)))$$ where $i,j\in\{1,2,3...n\}$. This is as far as I can get.

Answer 1

It looks like this can be proven by induction. I haven't nailed everything down yet but the idea is that suppose the statement is true for n-1, we have the expansion $$S^{n-1}=1+(n-1)[\text{linked-cluster diagrams}]+(n-1)^2[\text{products of 2 linked-cluster diagrams}]+...$$ Now it can be checked that $$S^n=S\times S^{n-1}=(1+[\text{linked-cluster diagrams}]+...)(1+(n-1)[\text{linked-cluster diagrams}]+(n-1)^2[\text{products of 2 linked-cluster diagrams}]+...)$$ Hopefully the combinatorics works out so that $${[\text{products of m}][\text{product of n}]={n+m\choose m}[\text{products of m+n}]}^*$$
then it can be checked that the coefficient before $[\text{products of l]}$ is $$\sum^{l}_{i=0} {{l}\choose i}(n-1)^i=n^l$$ Q.E.D. But the Feynman diagram combinatorics scares me so I'm not sure how to prove that or if it's true.

Edit: This is probably wrong. The relation between the symmetry factors of the product of two diagrams and that of each diagram seems to be complicated, i.e, $$\bigg[\text{Diagram A}\bigg]\times\bigg[\text{Diagram B}\bigg]\ne\bigg[\text{Diagram A and Diagram B put side by side}\bigg]$$ because of the new symmetries arising from joining A and B.

Edit 2: It seems reasonable that these new symmetries might be exactly canceled from the overcounting of the identical partitions into (m,n) of Linked-clustered diagram? This looks promising heuristically as I can't think of a counter example. Group theory knowledge needed?

In summary, to show $^*$, for each diagram $D_i$ in [products of m+n], we can partition it to a product of m LCD (Linked-Clustered Diagram) $F_j$ and a product of n LCD $R_k$, which we can take from each of the sums from the LHS to form. However, if there are some identical LCDs in the $D_i$, assuming the rest are different, the symmetry factor will be some factor, say 1/r, times the product of the symmetry factors of $F_j$ and $R_k$ depending on how the identical LCDs are partitioned. But at the same time, we are overcounting the identical partitions by r times. And this concludes our argument.