Observe that we have $Q(\sqrt{2})=Q((\sqrt{2}+1)^n)$.


More generally, assume that $K$ is a finite extension of Q. Is there any $\alpha \in K$ such that $K=Q(\alpha^n)$ for every $n \in N$?



$Q(\sqrt{2})=Q((\sqrt{2}+1)^n)$
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 $K$-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 again a strict $K$-subspace of $L$ because $\sigma\_i(1)=\sigma\_j(1)=1$. Since the union of finitely many 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 $\alpha \mapsto \overline{\alpha}$.


