$Q(\sqrt{2})=Q((\sqrt{2}+1)^n)$ 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 $n \in N$?
 A: The answer for non-real quadratic extensions of ${\mathbb Q}$ is Yes. For most elements $\alpha\in K$, the number $\gamma:=\frac\alpha{\bar\alpha}\in K$ is of modulus $1$ but not a root of unity. The reason is that quadratic roots of unity are finitely many, and an equation $\bar\beta=x\beta$ determines a line in the lattice $K$.
The fact that $\gamma$ is not a root of unity ensures that for every $n\ge1$, 
$$\alpha^n=p_n\alpha+q_n,\qquad p_n,q_n\in{\mathbb Q},$$
and
$$p_n=\frac{\bar\alpha^{n}-\alpha^{n}}{\bar\alpha-\alpha}\neq0.$$
Therefore $K={\mathbb Q}(\alpha^n)$ for every $n\ge1$.
A: $\def\QQ{\mathbb{Q}}$This question seems unmotivated to me, but I might as well answer it: Yes, for any finite extension $K \supset \QQ$, there is some $\alpha \in K$ such that $\QQ(\alpha^n) = K$ for all nonzero integers $n$.
$\def\p{\mathfrak{p}}$Let $p$ be a prime which splits completely in $K$, say $p = \p_1 \p_2 \ldots \p_d$ with $d=[K:\QQ]$. We know that $p$ exists by Cebotarov. By the Chinese remainder theorem, there is some $\alpha$ such that the $d$ valuations $v_{\p_i}(\alpha)$ are all distinct. So, for any $n$, the $d$ valuations $v_{\p_i}(\alpha^n)$ are all distinct.
If $L$ is a proper subfield of $K$, then there is some pair $\p_i$ and $\p_j$ such that $\p_i \cap L = \p_j \cap L$. (Because $p$ can only break into at most $[L:\QQ]$ factors in $L$.) So, if $\alpha^n$ were in $L$, then $v_{\p_i}(\alpha^n)$ would equal $v_{\p_j}(\alpha^n)$. This contradiction shows that $\alpha^n$ is not contained in any subfield of $K$.
Remark: One can make the same argument work using the archimedean place, if it happens that $K$ is totally real. This is what is happening in the case of $\QQ(\sqrt{2})$.  
A: 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 $\alpha \mapsto \overline{\alpha}$.
A: Choose a prime $p$ that splits completely in $K$, and let
$\mathfrak{P}$ denote a prime in $\mathcal{O}_K$ above $p$. There
exists an integer $h$ such that $\mathfrak{P}^h$ is principal;
write $\mathfrak{P} = (\alpha)$.  I claim that
$K = \mathbf{Q}(\alpha^n)$ for all $n$. 
To see this, suppose that $\alpha^n$ lived inside some proper 
subfield $E$. It's easy to see that the ideal $\alpha^n \mathcal{O}_E$ is divisible by $\mathfrak{p} =
\mathfrak{P} \cap \mathcal{O}_E$. But, by construction, the prime
$\mathfrak{p}$ splits into $[K:E] > 1$ distinct primes in $K$, which
is incompatible with the factorization $(\alpha^n) = \mathfrak{P}^{nh}$
in $\mathcal{O}_K$.
