jmc

We let $e_1={\zeta }_1+{\zeta }_2+{\zeta }_3,e_2={\zeta }_1{\zeta }_2+{\zeta }_2{\zeta }_3+{\zeta }_3{\zeta }_1,e_3={\zeta }_1{\zeta }_2{\zeta }_3$ (the elementary symmetric sums). Then, we can rewrite the above equations as$${\zeta }_1+{\zeta }_2+{\zeta }_3=e_1=1$$$${\zeta }_1^2+{\zeta }_2^2+{\zeta }_3^2=e_1^2-2e_2=3$$from where it follows that $e_2=-1$. The third equation can be factored as$$\begin{gathered} 7={\zeta }_1^3+{\zeta }_2^3+{\zeta }_3^3=({\zeta }_1+{\zeta }_2+{\zeta }_3)({\zeta }_1^2+{\zeta }_2^2+{\zeta }_3^2-{\zeta }_1{\zeta }_2-{\zeta }_2{\zeta }_3-{\zeta }_3{\zeta }_1)+3{\zeta }_1{\zeta }_2{\zeta }_3 \\ =e_1^3-3e_1{e}_2+3e_3, \end{gathered}$$from where it follows that $e_3=1$. Thus, applying Vieta's formulas backwards, ${\zeta }_1,{\zeta }_2,$ and ${\zeta }_3$ are the roots of the polynomial$$x^3-x^2-x-1=0⟺x^3=x^2+x+1$$Let $s_n={\zeta }_1^n+{\zeta }_2^n+{\zeta }_3^n$ (the power sums). Then from $(1)$, we have the recursion $s_{n+3}=s_{n+2}+s_{n+1}+s_n$. It follows that $s_4=7+3+1=11,s_5=21,s_6=39,s_7=\boxed{71}$.