How to show $\mathbb{Q}(\alpha^{4})=\mathbb{Q}(\alpha)$? From Berkeley Problems in Mathematics, Spring 1999, Problem 17.
Let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree $n\ge 3$. Let $L$ be the splitting field of $f$, and let $\alpha\in L$ be a zero of $f$. Give that $[L:\mathbb{Q}]=n!$, prove that $\mathbb{Q}(\alpha^{4})=\mathbb{Q}(\alpha)$. 
(BELOW IS NONSENSE, SHOULD BE IGNORED)
Following Gerry Myerson's advice, since $L$ is the splitting field of $f$, $L$ must be Galois over $\mathbb{Q}$. Thus the Galois group for $[L:\mathbb{Q}]=S_{n}$. Gerry now asserts that either $\mathbb{Q}(\alpha^{4})$ is of degree 2 over rationals or it must be $\mathbb{Q}$. But assuming this one can rule out the case $\mathbb{Q}$, since this would imply $\alpha$ is the root of a 4th or 2nd degree polynomial over $\mathbb{Q}$. But we know $n\ge 3$. So $\alpha$ must be the root of a 4th degree polynomial. We know $S_{n}$ has normal subgroups $A_{n}$ when $n\not=4$ and $V,A_{4}$ when $n=4$. I do not know how to proceed any further. 
Now assuming $\mathbb{Q}(\alpha^{4})$ is a degree 2 abelian extension over $\mathbb{Q}$. We should have a chain of normal extensions $\mathbb{Q}(\alpha)\supset \mathbb{Q}(\alpha^{2})\supset \mathbb{Q}$. This would imply $S_{n}$ has a normal subgroup which has a normal subgroup. But we know $A_{n}$ for $n\ge 5$ are simple. So the only possibility is $\mathbb{Q}(\alpha)$ has a Galois group isomorphic to $V$. In this case $n=4$ as well. I can only proceed to here. 
 A: Look at an example. Let $f(x)=x^3-2$. You should be able to work out the splitting field, and see that it has degree 6 over the rationals. If you have any problem doing this, come back and let us know where you get stuck. 
EDIT: much of what I wrote a few hours ago was not right. Let me try again. 
$L$ is normal over the rationals, over $K={\bf Q}(\alpha^4)$, and over $E={\bf Q}(\alpha)$. The group of $L$ over the rationals is $S_n$, the group of $L$ over $E$ is $S_{n-1}$, so the group of $L$ over $K$ is a subgroup $H$ of $S_n$ containing $S_{n-1}$. We're trying to prove that $H=S_{n-1}$. 
We can rule out $H=S_n$, as follows. If $H=S_n$, then $\alpha^4=q$ is rational, and the minimal polynomial for $\alpha$ over the rationals is $x^4-q$ (or some factor of that polynomial), and we're not talking about a polynomial with Galois group $S_n$. 
So all we have to show is that there is no proper subgroup of $S_n$ properly containing $S_{n-1}$, and we're done. And, hey, there's a proof on m.se, so we're done.  
I think what follows can safely be ignored.   
Back to the ${\bf Q}(\alpha^4)$ question: if ${\bf Q}(\alpha^4)$ is a proper subfield of ${\bf Q}(\alpha)$, then it's either the rationals or degree 2 over the rationals. It's easy to rule out the first case (details left to you - I have to teach a class in a few minutes). In the second case, it's normal over the rationals, and ${\bf Q}(\alpha)$ is normal over it, and that says something about normal subgroups of the Galois group of $L$ which don't fit with that group being the symmetric group, $S_n$ (that group doesn't have a lot of normal subgroups). 
I know I've left a lot out. I hope it's of some use. But if you haven't done the Galois Theory, my answer won't do much for you. 
A: I wish to give a solution that do not require to know anything about the normal subgroups of $S_n$:
First note $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^{4})]\leq4$ since
$\alpha$ is a root of $x^{4}-\alpha^{4}\in\mathbb{Q}(\alpha^{4})[x]$.
Now there is need to divide to cases: If the degree of the extension is $1$ then you are done, we need
to show that it can't be $2,3,4$.
If the degree of the extension is $3$ then since $\mathbb{Q}(\alpha^{2})$
is a subextension we have $$3=[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^{4})]=[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^{2})][\mathbb{Q}(\alpha^{2}):\mathbb{Q}(\alpha^{4})]$$
hence one of $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^{2})],[\mathbb{Q}(\alpha^{2}):\mathbb{Q}(\alpha^{4})]$
is of degree $3$ which is clearly a contradiction since, for example,
$\alpha$ is a root of $x^{2}-\alpha^{2}\in\mathbb{Q}(\alpha^{2})$.
if $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^{2})]=2$ then there is
$g(x)=x^{2}+ax+b\in\mathbb{Q}(\alpha^{2})$ s.t. $g(\alpha)=0$, but
then $g|f$ in $\mathbb{Q}(\alpha^{2})[x]\implies f=gh$ where $h\in\mathbb{Q}(\alpha^{2})[x],\deg(h)=\deg(f)-\deg(g)=n-2$.
Now, $f$ is irreducible (over $\mathbb{Q})$ hence $[\mathbb{Q}(\alpha):\mathbb{Q}]=\deg(f)=n$
so $[\mathbb{Q}(\alpha^{2}):\mathbb{Q}]=\frac{n}{2}$.
Denote $M$ as the splitting field of $h$ over $\mathbb{Q}(\alpha^{2})$,
then $[M:\mathbb{Q}(\alpha^{2})]\leq(n-2)!$, also note that since
$[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^{2})]=2$ then $[M(\alpha):\mathbb{Q}(\alpha^{2})]\leq2(n-2)!$
and from the previews line $[M(\alpha):\mathbb{Q}]\leq2(n-2)!\frac{n}{2}=(n-2)!n<n!$
That is: $M(\alpha)$ is a field over $\mathbb{Q}$ containing all
the roots of $f$ and is of degree $<n!$, and this is a contradiction.
In the same manner you get a contradictions to the rest of the cases
which are similar, I leave it out to you to fill the rest of the details
(I am in a hurry to get to class too). 
