Let $M=E[e^{t\cdot X}]$ be the moment-generating function of the random vector $X$ in $\mathbb{R}^n$. Then is it true that $$ E[(X -EX)^s] = \left. \frac{\partial \ln{M}}{\partial t^s} \right|_{t=0} $$ where $s \in \mathbb{N}^n$ and $$ X^s = X_1^{s_1}\cdots X_n^{s_n}, \qquad \frac{\partial}{\partial t^s} =\frac{\partial}{\partial t_1^{s_1}} \cdots \frac{\partial}{\partial t_n^{s_n}} $$
I only checked this for $|s| = s_1+\cdots+s_n \le 3$, but it seems to be true.
EDIT: If $|s|=1$, then the formula doesn't hold. However, it seems to hold for at least $|s|=2,3$
EDIT 2: Realized that the formula doesn't hold for $|s|=4$. It seems that the $|s|=2,3$ cases were indeed a fluke
EDIT 3: I realized that what I'm essentially asking is the cumulant expansion. The "fluke" is actually documented on wikipedia.
Let $n=1$. Then $\frac d {dt} \ln M(t)=\frac {M'(t)} {M(t)}$. At $t=0$ this has the value $EX$ nor $E(X-EX)$.