Why is the sum of all external angles in a convex polygon $360^\circ$ and not $720^\circ$? 
Why is the sum of all external angles in a convex polygon $360^\circ$? 

From my understanding, for each vertex in a convex polygon, there exist exactly $2$ exterior angles corresponding to it, which are both equal, vertically opposite, and add up to $180^\circ$ with the interior angle. If we take as true that sum of interior angles in a triangle is $(n-2)180^\circ$ degrees, then $$\sum_i 2\cdot (180^\circ-\alpha_i) = n\cdot 360^\circ - (n-2)\cdot 360^\circ = 720^\circ.$$ Am I missing something here? 
 A: As the comment says, there are two equal exterior angles at each vertex, one on the left of the vertex and one on the right. When we say that "the sum of the exterior angles is 360°", we mean that the sum of the left-side angles is 360° and that the sum of the right-side angles is 360°, not that the sum of the two sets together is 360°.
A: I recommend you look at Geometry by Jurgensen under the angles of a polygon section.
Draw a picture, it will help. Start out with pentagon to get some intuition. Now separate the inside of the polygon into non-overlapping triangles, and observe how you found the total sum of the interior angles of the polygon, and then of the sum of the exterior angles of it.
Basically, given a convex polygon, we can form a linear pair -- a pair of supplementary angles , at each vertex.
Taking all $n$ vertices of the polygon into account, we have $n \cdot 180 $ degrees.
The polygon can be partitioned/separated into exactly $n-2$ triangles.
So, the sum of all the interior angles of the polygon is $$(n-2) \cdot 180 \text{ degrees} $$
Hence, the sum of all the external angles is the difference
$$ n \cdot 180 - (n-2) \cdot 180 = 2\cdot 180  = 360 \text{ degrees} $$
