Solving a problem using Cauchy's residue theorem, is there more to it? Let $z_1,...z_n$ be distinct complex numbers. Let $C$ be a circle around $z_1$ such that no other $z_j$ is in $C$ for $j>1$. Let $$f(z) = (z-z_1)(z-z_2)...(z-z_n)$$ Find $\oint_{C}{\dfrac{\mathrm{d}z}{f(z)}}$. 
Attempt: Using Cauchy's Residue Theorem, and $$g(z) = \dfrac{1}{f(z)}$$ we see that $g(z)$ is analytic inside and on $C$ (assuming no $z_j$ is on $C$ for $j>1$) except for $z=z_1$ $$\oint_{C}{\dfrac{\mathrm{d}z}{f(z)}}=\oint_{C}{g(z)}\ \mathrm{d}z=2\pi\textbf{i}\ \phi(z_1)$$ where$$\phi(z) = \dfrac{1}{(z-z_2)(z-z_3)...(z-z_n)}$$ $$\therefore \phi(z_1) = \dfrac{1}{(z_1-z_2)...(z_1-z_n)}$$
Any faults with this? Or something that I should add?
 A: Cauchy's Residue theorem states:

Let $C$ be a simple closed contour, oriented in a positive direction.
  If a function is analytic inside and on $C$ except for a finite number
  of singular points denoted by $z_k\ \forall\ k \in \{1,2..n\}$ then:
  $$\oint_C f(z)\ \mathrm{d}z = 2\pi \textbf{i}\sum_{k=1}^{n}\mathrm{Res}_{z=z_k}f(z)$$

From Brown and Churchill, Complex Variables and applications.
From what you've written:

Let $C$ be a circle around $z_1$ such that no other $z_j$ is in $C$ for $j>1$.

I understand that the Contour $C$ encloses $z=z_1$ and that all other points, $z_j$ with $j>1$ do not reside in $C$. Therefore we need to find the residue at $z=z_1$ and only at that point. Because it is the only point in the area enclosed by the Contour for which the function "blows up". For that we can use your method:
Let $g(z)=\frac{1}{f(z)}$ for which we have:
$$\oint_C g(z)\ \mathrm{d}z = 2\pi \textbf{i}\ \mathrm{Res}_{z=z_1}\frac{1}{f(z)}$$
To find the residue, apply the theorem:

An isolated singular point $z_0$ of some
  function $f$ is a pole of order $m$ iff $f(z)$ can be expressed
  thusly: $$f(z)=\frac{\phi{(z)}}{(z-z_0)^m}$$

In our case $z_0$ is $z_1$, which is isolated by virtue of the contour we have chosen to enclose it. So our isolated point is $z_1$ and the pole is simple, i.e. $m=1$
$$g(z) = \frac{1}{f(z)} = \frac{\phi(z)}{(z-z_1)}$$
From here we see that we can indeed express our function in the form indicated and that $\phi(z)$ is in fact $\frac{1}{(z-z_2)...(z-z_n)}$.
So the Residue is simply:
$$\mathrm{Res}_{z=z_1}g(z) = \phi(z_1) = \frac{1}{(z_1-z_2)...(z_1-z_n)}$$
So:
$$\oint_C f(z)\ \mathrm{d}z = 2\pi \textbf{i}\ \phi(z_1) = \frac{2\pi \textbf{i}}{{(z_1-z_2)...(z_1-z_n)}}$$
Hope this helps
