complex numbers - $z=e^{2pi i/5}$ solves $1+z+z^2+z^3+z^4=0$

What is the best way to verify that $$1+z+z^2+z^3+z^4=0$$ given $z=e^{2\pi i/5}$?

I tried using Euler's formula before substituting this in, but the work got messy real fast.

Answer

Clearly(?) $z^5=1$ and $z\ne 1$. Also, $$z^ 5-1=(z-1)(1+z+z^2+z^3+z^4)$$