True. For three terms $1+i-i=1$, all of which are $4^{th}$-root of $1$. For two terms you can also write $-\frac{1+\sqrt{3}i}{2}-\frac{1-\sqrt{3}i}{2}=-1$, all of which are $6^{th}$-root of $1$. And so on and so on…
Edit: Now that you edited the question and $p$ becomes prime, there is a more general answer. As Douglas pointed out the key word here is cyclotomic polynomials $P_p(z)=\sum_{n=0}^{p-1}z^n$, whose roots are precisely the $p^{th}$-root of unity except $1$. In that case
$1+P(z_j)=1$
gives you such a relation as long as $z_j\neq 1$.