Consider a hexagon and all its exterior angles. Let $S$ be the sum of the measures of the exterior angles.

Notice that an interior angle and its exterior angle form a linear pair. Therefore, the sum of their measures is equal to $180^{\circ }.$ $$\{\begin{array}{l}m\angle 1+m\angle A=180^{\circ }\\ m\angle 2+m\angle B=180^{\circ }\\ m\angle 3+m\angle C=180^{\circ }\\ m\angle 4+m\angle D=180^{\circ }\\ m\angle 5+m\angle E=180^{\circ }\\ m\angle 6+m\angle F=180^{\circ }\end{array}$$ By the Polygon Interior Angles Theorem, the sum of the measures of the interior angles of a polygon with $n$ sides is $(n-2)180^{\circ }.$ With this information, add all the equations in the system above. $$\begin{array}{rl}m\angle 1+m\angle A& =180^{\circ }\\ m\angle 2+m\angle B& =180^{\circ }\\ m\angle 3+m\angle C& =180^{\circ }\\ +m\angle 4+m\angle D& =180^{\circ }\\ m\angle 5+m\angle E& =180^{\circ }\\ m\angle 6+m\angle F& =180^{\circ }\\ S+(6-2)\cdot 180^{\circ }& =6\cdot 180^{\circ }\end{array}$$ Finally, by solving the last equation for $S,$ the desired result is obtained.

Although the proof above used a hexagon, the same reasoning can be applied no matter the number of sides of the polygon.