First recall for a nontrivial $p$-th root of unity $\zeta$, that is $\zeta \neq 1$, you have that $\zeta^j$ with $j=0, \dots, p-1$ is the set of all $p$-th roots of unity. So if there are roots of unity summing to $0$, this means $\sum_{j \in J} \zeta^j = 0$ for some subset $J$ of $\{0, \dots, p-1\}$.
This means $\zeta$ is a root of $\sum_{j \in J} X^j$. Yet the minimal polynomial of $\zeta$ is $\sum_{j = 0}^{p-1} X^j$. The minimal polynomial of an algebraic number (over the rationals) is the monic polynomial (with rational coefficients) of lowest degree that has this number as a root; equivalently, it is the irreducible monic polynomial having this number as a root. Each polynomial (with rational coefficients) having that algebraic number as a root must be a multiple of the minimal polynomial.
Since $\sum_{j \in J} X^j$ cannot be a multiple of $\sum_{j = 0}^{p-1} X^j$, except if they are equal, the claim follows.
What remains to check is that $\sum_{j = 0}^{p-1} X^j$ is the minimal polynomial.
Since this is equal to $(X^p-1)/(X-1)$ it is clear that $\zeta$ is a root. It remains to show that it is irreducible. This can be done by applying Eisenstein's criterion to this polynomial shifted by $1$; see page 2 of Section 16 of some Algebra notes by Paul Garrett
