Observe that $M(t)=E[e^{tX}]=E[1+X+\frac{(tX)^2}{2!}+\frac{(tX)^3}{3!}+\dots]$ since $e^a=1+a+\frac{a^2}{2!}+\frac{a^3}{3!}+\dots$. If the random variable $e^{|tX|}$ has a finite expected value then we can switch the sum and the expected valued to obtain:
$$M(t)=E[\sum_{n=0}^{\infty}\frac{(tX)^n}{n!}]=\sum_{n=0}^{\infty}\frac{E[(tX)^n]}{n!}=\sum_{n=0}^{\infty}t^n\frac{E[X^n]}{n!}$$
This implies that the $n^{th}$ derivative of $M(t)$ evaluated at $t=0$ is exactly $E[X^n]$