Is it possible that $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^{n})$ for all $n>1$? Is it possible that $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^{n})$ for all $n>1$ when $\mathbb{Q}(\alpha)$ is a $p$th degree Galois extension of $\mathbb{Q}$? 
($p$ is prime)
I got stuck with this problem while trying to construct polynomials whose Galois group is cyclic group of order $p$.
Edit: Okay, I got two nice answers for this question but to fulfill my original purpose(constructing polynomials with cyclic Galois group) I realized that I should ask for all primes $p$ if there is any such $\alpha$(depending on $p$) such that the above condition is satisfied. If the answer is no (i.e. there are  primes $p$ for which there is no $\alpha$ such that $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^{n})$ for all $n>1$) then I succeed up to certain stage.

$\alpha=1+\sqrt 2$ (for $p=2$) should do the trick. By the binomial theorem, $(1+\sqrt 2)^n$ is always $a_n+b_n\sqrt2$ for some positive integer $b_n$.
