What is needed to justify taking the derivative of a geometric series? Of course, if $|p| < 1$,
$$\sum_{n=0}^{\infty}p^n=\dfrac{1}{1-p}\text{.}$$
One fact that is particularly useful in probability and financial mathematics is taking the derivative of both sides leads to 
$$\sum_{n=1}^{\infty}np^{n-1}=\dfrac{1}{(1-p)^2}\text{.}$$
What is needed to justify doing this?
I was reading the Real Analysis text by Bartle yesterday, and my recollection is that if $(f_n) \to f$ converges at some $x \in S$, $S \subseteq \mathbb{R}$ bounded, 
and if $(f^{\prime}_n) \to g$ uniformly, then $(f_n) \to f$ uniformly and $f^{\prime} = g$ (I imagine this is true only over $S$).
Define $f_n: (0, 1) \to \mathbb{R}$ by $f_n(p) = p^n$. Obviously, $(f_n) \to 0$... oh wait, I realized I'm not working with a sequence of functions, but a series (as I'm typing this). So my method is likely invalid.
I'm not asking for a complete answer (go ahead and give one if you'd like), but a list of theorems or a sketch that would assist in proving this problem. This is not homework.
 A: In general
Let $f_n:\mathbb R\longrightarrow \mathbb R$ be a sequence of differentiable functions s.t. 
1) $(f_n)_n$ converge pointwise to $f$,
2) $(f_n')_n$ converge uniformly to $g$.
Then $f'=g$, i.e.
$$\left(\lim_{n\to \infty }f_n(x)\right)'=\lim_{n\to \infty }f_n'(x).$$
In other word, you can permute the limit and derivative.

On your problem
We suppose that the convergence radius of the series $$\sum_{k=0}^\infty a_nx^n$$
is $\mathcal R$. By a theorem of your course, you know that the radius of convergence of the series $$\sum_{k=1}^\infty na_nx^{n-1}$$ is also $\mathcal R$. Let $f_n(x)=\sum_{k=1}^n a_nx^n$. By an other theorem of your course, $f_n$ converge uniformly on all compact $K\subset ]-\mathcal R,\mathcal R[$, and $f_n'$ also. Therefore, for all $|x|<\mathcal R$, you have that $$\left(\sum_{k=0}^\infty a_nx^n\right)'=\left(\lim_{n\to \infty }\sum_{k=0}^\infty a_nx^n\right)'=\lim_{n\to \infty }\left(\sum_{k=0}^na_nx^n\right)'=\lim_{n\to \infty }\sum_{k=1}^nna_nx^{n-1}=\sum_{k=1}^\infty na_nx^{n-1}.$$

Conclusion
You can always differentiate series terms by terms when $|x|<\mathcal R.$
A: You can fix $k$ and consider series of the form
$$f_k(p)=\sum_{n=0}^kp^n=\frac{1-p^{k+1}}{1-p}$$
Then $f_k$ is differentiable and
$$f_k'(p)=\sum_{n=1}^knp^{n-1}\tag{$*$}$$
The series $\sum_{n=1}^\infty np^{n-1}$ converges uniformly on sets of the form $\left\{p:|p|\leq r\right\}$, where $r<1$ by Weierstrass' test, or you can simply use the last term in $(*)$ above to obtain the same.
By by the result you stated, the limit function $f(p)=\sum_{n=0}^\infty p^n=\frac{1}{1-p}$ is differentiable on sets of the form $\left\{p:|p|\leq r\right\}$ for $r<1$, and so it is differentiable on $\left\{p:|p|<1\right\}$. From usual calculus and by comparing it with ($*$),
$$\frac{1}{(p-1)^2}=f'(p)=\lim_k f_k'(p)=\lim_k\sum_{n=0}^knp^{n-1}=\sum_{n=0}^\infty np^{n-1}$$
