First observe that the star may be separated into a single $n$-gon and $n$ triangles.
\begin{align*} (n-2) \cdot 180 &= \text{ sum of interior angles of the $n$-gon } \\ n \cdot 180 &= \text{sum of interior angles of all the $n$ triangles} \\ \end{align*}
The interior angles of the $n$ triangles not part of the tips may be labeled with $x_1, .... , x_n$ , using them twice where vertical angles arise.
$$ \sum_{j=1}^n 2x_j = \text{ sum of all angles of the $n$ triangles not part of the tips of the star}$$
Now see that each vertex of the on the the perimeter of the star that isn't a tip "represents" $360$ degrees. There are $n$ of these, so in total there is
$$ n \cdot 360 = 2n \cdot 180 \text{ degrees } $$ accounting for all of them.
If we exploit the vertical angles corresponding to the interior angles of the $n$-gon, we obtain
$$ \left[ \sum_{j=1}^n 2x_j \right] + 2(n-2) \cdot 180 = 2n \cdot 180 $$
Then $\sum_{j=1}^n 2x_j = 4 \cdot 180 $. To obtain the sum of the angles in the corners of the star, we simply do
$$ n \cdot 180 - 4 \cdot 180 = 180 \left[ n-4 \right], \, \text{ for } n \geq 5 $$