Complex Integral on Wikipedia Let $f(z)=\sum_{k=0}^\infty (-1)^k(z-1)^k$ and $D$ a disk of radius $r$ centered at $a$ with boundary $\partial D$. 
$\frac{f^{(k)}(a)}{k!}
=\frac{1}{2\pi i}\int_{\partial D}\frac{f(s)}{(s-a)^{k+1}}ds\\
=\frac{1}{2\pi i}\int_{\partial D}\frac{\sum_{n=0}^\infty (-1)^n(s-1)^n}{(s-a)^{k+1}}ds\\
=\frac{1}{2\pi i}\sum_{n=0}^\infty (-1)^n \int_{\partial D}\frac{(s-1)^n}{(s-a)^{k+1}}ds\\
=\frac{1}{2\pi} \sum_{n=0}^\infty (-1)^n\int_0^{2\pi}\frac{(a-1+re^{it})^n}{(re^{it})^k}dt\\
=\frac{1}{2\pi} \sum_{n=0}^\infty (-1)^n\int_0^{2\pi}\frac{\sum_{m=0}^n{n\choose m}(a-1)^{n-m}(re^{it})^m}{(re^{it})^k}dt\\
=(-1)^ka^{-k-1}
$
This computation can be found here. I am able to follow everything up until the last simplification. Essentially, this is just using the Cauchy Integral Formula and then plugging in the Taylor Series for $f(s)$ into the integral. I understand parameterizing $\partial D$ with $a+re^{it}$ for $0\leq t<2\pi$ since it is a disk and using the binomials expansion to get the other sum inside the integral, but I can't see how they went from the infinite sum and integral to just $(-1)^ka^{-k-1}$. I would assume they have used the Cauchy Residue Theorem in some way  to avoid taking the integral explicitly, but I am unable to see exactly what they've done. Any hints or explanations are much appreciated!
 A: You can actually compute the integral quite explicitly.  Pulling all the summations and constant factors outside the integral, you get
$$\frac{1}{2\pi} \sum_{n=0}^\infty (-1)^n\sum_{m=0}^n{n\choose m}(a-1)^{n-m} r^{m-k}\int_0^{2\pi}e^{i(m-k)t}dt.$$  That integral can now be evaluated explicitly by elementary calculus: if $m-k\neq 0$ you get $\frac{1}{i(m-k)}(e^{i(m-k)\cdot 2\pi}-e^{i(m-k)\cdot 0})=0$, and if $m-k=0$ you get $2\pi$.  So all the terms in the sum go away except when $m=k$, and you're left with
$$\sum_{n=k}^\infty (-1)^n{n\choose k}(a-1)^{n-k}=(-1)^k\sum_{j=0}^\infty\binom{j+k}{k}(1-a)^j$$ (in the last step I reindexed the sum with $j=n-k$ and rearranged the factors of $-1$ a little).  You can now recognize this last sum as exactly the Taylor series for $g(x)=\frac{1}{(1-x)^{k+1}}$ evaluated at $x=1-a$, and so this is just $$(-1)^k\frac{1}{a^{k+1}}=(-1)^ka^{-k-1}.$$

That said, this whole exercise seems rather silly: if you're going to use the Taylor series for $\frac{1}{(1-x)^{k+1}}$ at the end, why not just explicitly compute the derivative $f^{(k)}(a)$ using the fact that $f(z)=1/z$, which amounts to essentially the same thing?  I don't think there's any deeper approach you're missing here; this is just a poorly written example on Wikipedia.  The point of the example is that even assuming $f$ is defined only on the unit disk, if you use the higher derivatives of $f$ at $a$ to write out the Taylor series for $f$ centered around $a$, you get a series which you can explicitly see has a disk of convergence that extends beyond the original domain.  Exactly how you compute those higher derivatives is not an important point.
