Proof

By the Maclaurin series expansion of \(\rm{e}^{Xt}\), we can expand \(M_{X}(t)\) as follows:

\begin{eqnarray*}M_{X}(t)&=&E[\rm{e}^{Xt}]\\&=&E[1+\frac{X}{1!}t+\frac{X^{2}}{2!}t^{2}+\cdots +\frac{X^{n}}{n!}t^{n}+\cdots ]\\&=&E[1]+\frac{E[X]}{1!}t+\frac{E[X^{2}]}{2!}t^{2}+\cdots +\frac{E[X^{n}]}{n!}t^{n}+\cdots\end{eqnarray*}

From the expansion, we can express the moments \(E[X^{k}]\), the coefficient of \(\frac{t^{k}}{k!}\), as below.

$$E[X^{k}]= M^{(k)}_{X}(0)=\left[\frac{d^{k}}{dt^{k}}M_{X}(t)\right]_{t=0}$$

where \(M^{(k)}_{X}(0)\) is the k-th derivation of \(M_{X}(t)\) respect to \(t\) and evaluated at \(t=0\).