In a more general setting, the following is true : let $K$ be an infinite field and $L/K$ be a finite separable extension whose Galois closure $M$ contains only finitely many roots of unity (this assumption is true for number fields). Then there exists $\alpha \in L$ such that $L=K(\alpha^n)$ for every $n \geq 1$.
Proof : let $d=[L:K]$ and $\sigma_1,\ldots,\sigma_d$ be the distinct $K$-embeddings of $L$ into $M$. Given $\alpha \in L$, we have $L=K(\alpha)$ if and only if $\sigma_1(\alpha),\ldots,\sigma_d(\alpha)$ are pairwise distinct. If $i \neq j$ then the equation $\sigma_i(\alpha)=\sigma_j(\alpha)$ determines a strict $K$-hyperplane K$-subspace of$L$. Moreover, the equation$\sigma_i(\alpha^n)=\sigma_j(\alpha^n)$is equivalent to$\sigma_i(\alpha)= \zeta \cdot \sigma_j(\alpha)$for some$\zeta \in \mu_n(M)$. The set of$\alpha$satisfying the last condition is either empty or again a strict$K$-hyperplane K$-subspace of $L$. L$because$\sigma_i(1)=\sigma_j(1)=1$. Since the union of finitely many hyperplanes strict subspaces of$L$cannot be equal to$L$, the result follows. EDIT. Note that this proof is based on the same idea as in Denis's answer : if$K$is a quadratic field, the embeddings of$K$are just the identity map and the map$\alpha \mapsto \overline{\alpha}$. 2 added 204 characters in body In a more general setting, the following is true : let$K$be an infinite field and$L/K$be a finite separable extension whose Galois closure$M$contains only finitely many roots of unity (this assumption is true for number fields). Then there exists$\alpha \in L$such that$L=K(\alpha^n)$for every$n \geq 1$. Proof : let$d=[L:K]$and$\sigma_1,\ldots,\sigma_d$be the distinct$K$-embeddings of$L$into$M$. Given$\alpha \in L$, we have$L=K(\alpha)$if and only if$\sigma_1(\alpha),\ldots,\sigma_d(\alpha)$are pairwise distinct. If$i \neq j$then the equation$\sigma_i(\alpha)=\sigma_j(\alpha)$determines a$K$-hyperplane of$L$. Moreover, the equation$\sigma_i(\alpha^n)=\sigma_j(\alpha^n)$is equivalent to$\sigma_i(\alpha)= \zeta \cdot \sigma_j(\alpha)$for some$\zeta \in \mu_n(M)$. The set of$\alpha$satisfying the last condition is either empty or again a$K$-hyperplane of$L$. Since the union of finitely many hyperplanes of$L$cannot be equal to$L$, the result follows. EDIT. Note that this proof is based on the same idea as in Denis's answer : if$K$is a quadratic field, the embeddings of$K$are just the identity map and the map$\alpha \mapsto \overline{\alpha}$. 1 In a more general setting, the following is true : let$K$be an infinite field and$L/K$be a finite separable extension whose Galois closure$M$contains only finitely many roots of unity (this assumption is true for number fields). Then there exists$\alpha \in L$such that$L=K(\alpha^n)$for every$n \geq 1$. Proof : let$d=[L:K]$and$\sigma_1,\ldots,\sigma_d$be the distinct$K$-embeddings of$L$into$M$. Given$\alpha \in L$, we have$L=K(\alpha)$if and only if$\sigma_1(\alpha),\ldots,\sigma_d(\alpha)$are pairwise distinct. If$i \neq j$then the equation$\sigma_i(\alpha)=\sigma_j(\alpha)$determines a$K$-hyperplane of$L$. Moreover, the equation$\sigma_i(\alpha^n)=\sigma_j(\alpha^n)$is equivalent to$\sigma_i(\alpha)= \zeta \cdot \sigma_j(\alpha)$for some$\zeta \in \mu_n(M)$. The set of$\alpha$satisfying the last condition is either empty or again a$K$-hyperplane of$L$. Since the union of finitely many hyperplanes of$L$cannot be equal to$L\$, the result follows.