For which $k$ do the $k$th powers of the roots of a polynomial give a basis for a number field? Let $f \in \mathbb{Q}[x]$ of degreee $d$ be irreducible, with roots $\alpha_1,\ldots, \alpha_d$.  One particular basis for the field extension of $\mathbb{Q}$ obtained by adjoining the roots of $f$ is given by $\{\alpha_j\}_{j=1}^d$.  For which $k$ does the set $\{\alpha_j^k\}_{j=1}^d$ also provide a basis?
If this is in general a difficult question, could I at least show that I obtain a basis for infinitely many $k$?
Example: Consider the polynomial $x^2-2$.  The $k$th powers of the roots $\{\pm \sqrt{2}\}$ provide a basis for my field extension precisely when $k$ is odd.
Edit:  As Gerry Myerson points out, and as my example indicates, I am actually more interested in when the $k$th powers of the roots generate the extension over $\mathbb{Q}$.  Calling this a $\mathbb{Q}$-basis was erroneous.
 A: Let $L$ be a field of characteristic $0$ and $K$ a finite simple separable extension, with primitive element $\alpha \in K$. Let $E$ be the galois closure of $K$ over $L$. So we have a tower of fields $L \subset L(\alpha)=K \subset E$. Now if for some $k \in \mathbb N$ we have that $L(\alpha^k) \subsetneq K$ then there exists some $\sigma \in \mathrm{Gal}(E,L)$ fixing $\alpha^k$ that does not fix $\alpha$. Now $\sigma(\alpha^k)=\sigma(\alpha)^k=\alpha^k$ in particular $\sigma(\alpha)$ is a solution to $x^k-\alpha^k$ and thereby $\sigma(\alpha)=\zeta \alpha$ where $\zeta$ is a $k$-th root of unity. We see that there are finitely many automorphisms of $E$ over $L$ and since there are infinitely many $p$-th roots of unity, we have that $L(\alpha^p)=L(\alpha)$ for all but finitely many primes $p$. This then extends naturally to the case of $n$ generators by taking intersections. 
In the case that the base field is $L=\mathbb Q$ we can make this slightly more effective by noting that if $\varphi(p)>[\mathbb Q(\alpha): \mathbb Q]$ then $\alpha^p$ is still a generator. Which can then be extended to composite numbers by imposing the same constraint on the prime factors. But this doesn't necessarily tell the whole story. For most primitive generators I would expect that for every $k \in \mathbb N$ $\alpha^k$ generates the field. Implicit in my previous argument is the observation that $\alpha^k$ does not generate $L(\alpha)$ if and only if $\min_L(\alpha) \mid x^k-\alpha^k$. Which seems to me the exceptional case. 
