Integrating a power series proof 
So I began with using the hint and got to $f(x) - \sum_{n=?}^{\infty} a_nx^n$ But I have a question about the ? in the sum obviously. Because it's been differentiated, do we have to increase the index to $1$ or can we keep it at $n = 0$ since it's $x^n$ in the sum? 
As a follow-on if "$?$" does equal to 0 then clearly $g'(x) = 0$. But I'm not sure how this helps with the proof, as all this tells us is that there are $a,b \in (-R,R)$ s.t. $\frac{g(b)-g(a)}{b-a}=0$. Is this true for all $a,b \in (-R,R)$ implying that $g(x)=0$ for all $x$ thus finishing the proof?
 A: The index of the sum is just a notational decision; you can change the indexing, but be sure you know what you're doing, and that the result is the same.
For instance, before you differentiated, you had
$$
a_0 \dfrac{x^1}{1} + a_1 \dfrac{x^2}{2} + a_2 \dfrac{x^3}{3} \dots
$$ 
and afterward,
$$
a_0 x^0 + a_1 x^1 + a_2 x^2 \dots
$$
Is this the same as $\sum_{n=0} a_n x^n$?
For the other part of your question, "If $n=0$" only describes one term related to $g$; that is, it isn't a global condition or hypothesis that you need to explore.
Here's what I think you want to do here: suppose for contradiction that $F \neq \tilde F$. It is clear that $F(0) = \tilde F(0)$, so assume there is some other point $a$ such that $F(a) \neq \tilde F(a)$, then apply the mean value theorem to get a contradiction (you know that $g' \equiv 0$).
A: I always was annoyed by Calculus textbooks that teach to adjust the index when differentiating a power series. Because it's really unnecessary... if we're going to assume $0^0=1$, then why not go ahead and assume $0^{-p}=0$ too?
$$
\frac{d}{dx}\sum_{n=0}^{\infty} c_nx^n 
  = 
\sum_{n=0}^{\infty}c_n(nx^{n-1})
  =
c_n(0x^{-1}) +
\sum_{n=1}^{\infty}c_n(nx^{n-1})
  =
\sum_{n=1}^{\infty}c_n(nx^{n-1})
$$
In your case it doesn't even matter: the first term of $\sum_{n=0}^{\infty}a_n\frac{x^{n+1}}{n+1}$ has power $1$, not $0$. So the derivative is indeed $\sum_{n=0}^{\infty}a_nx^n$ (and $\sum_{n=1}^\infty$ would be incorrect).
