Proof

This is the special case of $E[X^k]=[\frac{d^k}{d{t}^k}{M}_X(t){]}_{t=0}$.

Actually, if we differentiate $M_X(t)=E[{\mathrm{e}}^{Xt}]$ respect to $t$, we have

$$M_X^{\prime }(t)=E[X{\mathrm{e}}^{Xt}]$$

and if we put $t=0$, we obtain

$$M_X^{\prime }(0)=E[X]$$