Total curvature of a closed polygonal curve Let $\tau$ be a closed polygonal curve in the plane, that is, $\tau$ consists of $n$ piecewise-linear segments between the vertices $v_1, \ldots, v_n, v_1$. The total curvature of $\tau$ is defined as the sum of the angles between the consecutive segments of the curve (i.e. the sum of the external angles). This picture from Wikipedia is self-explanatory:

It is known that the total curvature of such a curve is always an integer multiple of $2\pi$. But how to prove it?
I was able to prove the case in which $\tau$ defines something convex, i.e. a polygon. My idea was to triangulate the resulting polygon with vertex set $V$, edge set $E$ and face set $F$ (where F does not contain the outer face). Using then that in this particular case the total curvature $\kappa$ equals $\kappa = \pi(|V|-|F|)$ and applying Eulers formula gave the result $\kappa = 2\pi$.
What about non-convex "polygons"? What if the exterior angle actually can become "negative"? I am stuck and would greatly appreciate some help!
Thanks in advance
UPDATE: It would absolutely suffice to just consider the case of non-intersecting segments!
 A: That this total curvature, the signed sum of the external angles, is a multiple of $2\pi$ is obvious. Signed sum means that turning to the left is counted positive, turning to the right is counted negative. If your polygon is a Jordan curve, i.e., has no selfintersections, then Hopf's Umlaufsatz (turning tangent theorem, unfortunately not to be found on the english Wikpedia) says that the total curvature is $\pm2\pi$. Here is a copy of Hopf's paper:
http://www.math.harvard.edu/~knill/graphgeometry/hopf/
A: Let the vertices $\ v_1, v_2, \dots, v_n\ $ lie at the points $\ z_1, z_2, \dots, z_n\ $ in the complex plane, $\ z_0=z_n\ $, $\ z_{n+1}=z_1\ $, and $\ z_j-z_{j-1}= r_je^{i\theta_j}\ $ for $\ j=1,2, \dots, n+1\ $ with $\ r_j > 0\ $ and $\ 0\le \theta_j < 2\pi\ $.  If $\ \psi_j\ $ is the external angle at $\ v_j\ $ then $\ \psi_j \equiv \theta_{j+1}-\theta_j\ \left(\hspace{-0.5em}\mod 2\pi\right)$, and 
\begin{eqnarray}
\tau&=&\sum_\limits{j=1}^n\psi_j\\
&\equiv&\sum_\limits{j=1}^n \left(\theta_{j+1}-\theta_j\right)\left(\hspace{-1em}\mod 2\pi\right)\\
&\equiv& \theta_{n+1}-\theta_1\left(\hspace{-1em}\mod 2\pi\right)\ .
\end{eqnarray}
But $\ r_{n+1}e^{i\theta_{n+1}}=z_{n+1}-z_n=z_1-z_0=r_1e^{i\theta_1}\ $, so $\ \theta_{n+1}\equiv\theta_1\left(\hspace{-0.5em}\mod 2\pi\right)\ $ and $\ \tau\equiv 0\left(\hspace{-0.5em}\mod 2\pi\right)\ $.
