How to show $\mathbb{Q}(\alpha^{4})=\mathbb{Q}(\alpha)$?

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.

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).