(The following is not as elementary as the other answers, but it puts the given claim in a greater perspective.)
Using the proper sign conventions the claim is true even if the pentagon is non-convex, but still without self-intersections. Hopf's Umlaufsatz (sorry, there is no official English name for this theorem) says the following:
Theorem. The total curvature of a smooth Jordan curve $\gamma$ in the plane is $\pm 2\pi$.
In the case of a polygon the curvature is concentrated in the vertices, and the total curvature is the sum of the turning angles of the tangent vector at the vertices, or $a+b+c+d+e$ in your figure.