Let $X \sim \exp(\lambda)$. Prove that : $E[X^n] = n!/\lambda n$
The hint in (Fundamentals of probability) gives the hint use the Moment Generating Function, but I can't see how I should use that to prove that this is the respective expression that follows.
The MGF of the exponential distribution is given by \begin{align}M_X(t)&=\frac{λ}{λ-t}=\frac{1}{1-\frac{t}{λ}}=\sum_{n=0}^{\infty}\left(\frac{t}{λ}\right)^n=\sum_{n=0}^{\infty}\frac{t^n}{n!}\cdot\frac{n!}{λ^n}\end{align} by the formula of the geometric series, for $|t/λ|<1$. Now compare the above expression with the definition of the MGF $$M_X(t)=\mathbb E\left[e^{tX}\right]=\mathbb E\left[\sum_{n=0}^{\infty}\frac{t^nX^n}{n!}\right]=\sum_{n=0}^{\infty}\frac{t^n}{n!}\mathbb E[X^n]$$
The MGF is $m(t) = \frac{\lambda}{\lambda - t}$. We state, that the $n$-th derivative is $$ m^{(n)}(t) = n!\frac{\lambda}{(\lambda - t)^{n+1}}. $$ This is true for $n=0$, if it is true for $n$, we get \begin{align*} m^{(n+1)}(t) &= \frac{\lambda}{-(\lambda - t)}^{n+2}\cdot \bigl(-(n+1)\bigr) \\ &= (n+1)! \frac{\lambda}{(\lambda-t)^{n+2}} \end{align*} Therefore $$ \mathbf E[X^n] = m^{(n)}(0) = \frac{n!}{\lambda^{n}} $$
