Let $\phi$ be the characteristic function for random variable $X$. I know that if $E [|X|] < \infty$, then dominated convergence implies existence of the first derivative, and in particular, $\phi'(0) = i E[X]$. More generally if $E[|X|^k]<\infty$, then $\phi^{(k)}(0) = i^k E[X^k]$.
However, how can I argue the converse? Specifically, if I know $\phi$ is infinitely differentiable at $0$ and has a Taylor series $\sum_{k=0}^\infty \frac{t^k}{k!} \phi^{(k)}(0)$ that converges everywhere, how can I conclude that $\phi^{(k)}(0) = i^k E[X^k]$ and further that $E[|X|^k]<\infty$? I am not sure how to make the leap from existence of the derivative to claiming that it must be what we think it should be ($i^k E[X^k]$)...
Hint: First show that if $\phi$ has even order derivative then all moments exist upto that order of derivative. Here you have infinitely differentiable $\phi$ so obviously all even order derivatives exist and hence all moments exist.
By the way, the first part is a standard inclusion in probability textbooks.