Sums of central moments for a Gaussian distribution with mean zero and standard deviation one

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I want to know, whether the following calculations for a vector of variables $A_{i}$ of size N, drawn independently at random from a complex normal distribution, with mean zero and standard deviation one are correct:

$$\sum_{ij} \mathbb{E}(A_i \cdot A_j^*) = \sum_i \mathbb{E} (|A_i|^2) = N$$

$$\sum_{ijk} \mathbb{E}(A_i \cdot A_j^* \cdot A_k) = 0$$

$$\sum_{ijk} \mathbb{E}(A_i^* \cdot A_j \cdot A_k^*) = 0$$

I want to extend this to higher orders. I know, that the n-th central moment of a normal distribution is given by:

$$\hat{m}_{2n} = \mathbb{E}((A-\mathbb{E}(A))^{2n}) = \sigma^{2n} \cdot (2n-1)!!$$

$$\hat{m}_{2n+1} = 0$$

Is the following correct? I am especially interested in the intermediate step (So the first equation sign):

$$\sum_{ijkl} \mathbb{E}(A_i \cdot A_j^* \cdot A_k \cdot A_l^*) = \sum_{ik} \mathbb{E}(|A_i|^2) \cdot \mathbb{E}(|A_k|^2) \cdot (2n-1)!! = N^2 \cdot (2n-1)!! = 3 \cdot N^2$$

Finally for 2n variables:

$$\sum_{ijkl \cdots} \mathbb{E} (A_i \cdot A_j^* \cdot A_k \cdot A_l^* \cdots) = N^n \cdot (2n-1)!!$$