Say random variable X has a density function $ f(x)=1 $ when $0<x<1$. So this means $E[X^n]= \int_0^1 x^n.1 dx = \frac{1}{n+1}$.
At the same time we can get that the moment generation function $\phi(t)$ of random variable $X$ as $\phi(t)=\frac{e^t-1}{t}$
I know the $n^{th}$ derivation of $\phi(t)$ is $E[X^n]$ but how can I show $\phi^n(t)=\frac{1}{n+1}$
On
Hint:
If $\phi(t)$ denotes the moment generating function of $X$ then:$$\phi(t)=1+\frac{\mathbb EX.t}{1!}+\cdots\frac{\mathbb EX^n.t^n}{n!}+\cdots$$
If $\phi(t)=\frac{e^t-1}t$ then:$$\phi(t)=1+\frac{t}{2!}+\cdots+\frac{t^n}{(n+1)!}+\cdots$$
You cant show that, what you can show is that $\phi ^{(n)}(0) = \frac{1}{n+1}$ and in general you have $\mathbb{E}[X^k]= \phi ^{(k)}(0)$.
As written in the comments uyou could easily show that using the Taylor sum of $e^t$