Analytic Continuation of Complex Function I am triyng to solve the following problem in Brown and Churchill's complex variables textbook. 

Show that the function $f_2 (z) = 1/z^2$ ($z \neq 0$) is the analytic continuation of the function 
  \begin{align*}
f_1 (z) = \sum\limits_{n=0}^{\infty} (n+1)(z + 1)^n \ \ \ (|z+1| < 1)
\end{align*}
  into the domain consisting of all points in the $z$ plane except $z = 0$. 

As a first note, I am having difficulty mapping the definition of analytic continuation to this problem. The definition in the textbook is that if we have two domains, say $D_1$ and $D_2$, where some function $f_1$ is analytic on $D_1$, some function $f_2$ is analytic on $D_2$, and $f_1 (z) = f_2 (z)$ on $D_1 \cap D_2$, where this intersection is nonempty, then $f_2$ is the analytic continuation of $f_1$ into $D_2$. 
Assuming that I have not misstated that (please tell me if I have), we have: 
\begin{align*}
D_1 = \{z \in \mathbb{C} : |z + 1| < 1\}, \ \ \ D_2 = \{z \in \mathbb{C} : z \neq 0\}.
\end{align*}
So we have
\begin{align*}
D_1 \cap D_2 = \{z \in \mathbb{C} : |z + 1| < 1 \text{ and } z \neq 0\}.
\end{align*}
From here, I am stuck. I know I need to prove that $\frac{1}{z^2} = \sum\limits_{n=0}^{\infty} (n+1)(z + 1)^n$ for any $z \in D_1 \cap D_2$. I don't know if I should try to demonstrate that the moduli are equal or expand $\frac{1}{z^2}$ in a power series and hope that these results will match, subject to the given constraint. 
Any help would be greatly appreciated. 
EDIT: I do not believe this question is a duplicate. I looked through the link below, and it does not address this problem, nor does it seem to deal with concepts in complex analysis. 

We need to show that $1/z^2$ is the analytic continuation of $S$, so amongst other things we need to show they are equivalent on $D_1\cap D_2$.
Splitting the sum,
\begin{eqnarray}
S&=&\sum_{n=0}^\infty(z+1)^n+\sum_{n=0}^\infty n(z+1)^n\\
&=&\frac{1}{1-(z+1)}+(1+z)\frac{d}{dz}\sum_{n=0}^\infty (1+z)^n\\
&=&-\frac{1}{z}-(1+z)\frac{d}{dz}\frac{1}{z}\\
&=&-\frac{1}{z}+\frac{1+z}{z^2}\\
&=&\frac{1}{z^2}.
\end{eqnarray}
Since $S$ and $1/z^2$ are both analytic functions in the domains $D_1$ and $D_2$ respectively, and the intersection $D_1\cap D_2\neq\emptyset$, and furthermore $S=1/z^2$ on $D_1\cap D_2$, as shown above, then $1/z^2$ is the analytic continuation of $S$ to $\mathbb{C}\setminus\{0\}$, and vice-versa. Furthermore, this analytic continuation from $S$ to $1/z^2$ is unique.
