Succinct Proof: All Pentagons Are Star Shaped 
Question: What is a succinct proof that all pentagons are star shaped?

In case the term star shaped (or star convex) is unfamiliar or forgotten:
Definition Reminder: A subset $X$ of $\mathbb{R}^n$ is star shaped if there exists an $x \in X$ such that the line segment from $x$ to any point in $X$ is contained in $X$.  
This topic has arisen in the past in my class discussions around interior angle sums; specifically, for star shaped polygons in $\mathbb{R}^2$ we can find their sum of interior angles as follows:
By assumption, there is an interior point $x$ that can be connected to each of the $n$ vertices. Drawing in these line segments, we construct $n$ triangles; summing across all of their interior angles gives a total of $180n^\circ$, but this over-counts the angle sum for the polygon by the $360^\circ$ around $x$. Therefore, the sum of interior angles is $(180n - 360)^\circ = 180(n-2)^\circ$.
This formula for the sum of interior angles holds more generally (often shown by triangulating polygons) but the proof strategy above already fails for some (obviously concave) hexagons. 
For example:

The above depicted polygon is not star shaped.
Moreover, it is a fact that any polygon with five or fewer sides is star shaped. And so I re-paste:

Question: What is a succinct proof that all pentagons are star shaped?


Edit: Since it has come up as a counterexample (of sorts) for each of the first two responses, here is an example of a concave pentagon that may be worth examining in thinking through a proof.


A polygon must have at least three convex vertices (since each convex vertex turns by less than $\pi$ and in total the vertices must turn by $2\pi$). Thus there is at least one vertex that is between two convex vertices. This vertex sees all four other vertices, and hence the entire pentagon.
