If an element of a field is a primitive $7$th root, why are its powers also $7$th roots? If $F$ is a field containing an element $\alpha$ that is a primitive seventh root of $1$, why is it true that any power $\alpha^k$ is also a seventh root of $1$? Also, why is the set $ \\{1, \alpha, \alpha^2...\alpha^6\\}$ the full set of seventh roots in $F$?

If $\alpha \in F$ is a primitive seventh root of $1$, then it generates a cyclic subgroup of order $7$ inside $F$ consisting of $\\{1, \alpha, \alpha^{2}, \ldots, \alpha^{6}\\}$. By Lagrange, each of these elements has order dividing $7$, and since only $1$ has multiplicative order $1$, $\alpha, \alpha^{2}, \ldots, \alpha^{6}$ must each have order $7$, and therefore are also (primitive) seventh roots of $1$. In particular, each $\alpha^{k}$ for $k = 0, 1, \ldots, 6$ is distinct and is a root of $X^{7}-1$ over $F$. Since $F$ is a field, $X^{7} - 1$ has at most $7$ roots in $F$, so these powers of $\alpha$ must be all such roots.