What is a proof the following claim please?
If $X^1$ and $X^2$ are independent (the superscripts don't denote power), then all mixed cumulants involving $X^1$ and $X^2$ alone are zero.
Notation and clarification: here, I follow the notation of McCullagh (1987). Consider a random vector $X$ whose components are $X^1,\ldots,X^p$. Define the moments: $$ \kappa^i=E(X^i),\quad\kappa^{ij}=E(X^iX^j),\quad\kappa^{ijk}=E(X^iX^jX^k),\quad\ldots $$ and the moment generating function (for all $\xi\in\mathbb{R}^p$ such that the series below converge) $$ M_X(\xi)=1+\sum_i\xi_i\kappa^i+\sum_{i,j}\xi_i\xi_j\kappa^{ij}/2!+\sum_{i,j,k}\xi_i\xi_j\xi_k\kappa^{ijk}/3!+\cdots $$ and the cumulant generating function $K_X(\xi)=\log[M_X(\xi)]$. The cumulants $\kappa^i$, $\kappa^{i,j}$, $\kappa^{i,j,k}$, $\ldots$ (note the commas!) are then defined via $$ K_X(\xi)=1+\sum_i\xi_i\kappa^i+\sum_{i,j}\xi_i\xi_j\kappa^{i,j}/2!+\sum_{i,j,k}\xi_i\xi_j\xi_k\kappa^{i,j,k}/3!+\cdots $$
McCullagh claims that if $X^i$ and $X^j$ are independent, then $$ \kappa^{\underbrace{i,\ldots,i}_{r},\underbrace{j,\ldots,j}_{s}}=0,\quad\forall r,s\geq 1. $$ This is the claim I'm trying to understand.