The splitting field of $x^4-2$ is $\mathbb{Q}[i+2^{1/4}]$. I know the splitting field is generated by $2^{1/4}$ and $i$, I could show $\mathbb{Q} [ 2^{1/4}, i] = \mathbb{Q}[i+2^{1/4}]$ using some algebra. 
For the non trivial direction $\mathbb{Q} [ 2^{1/4}, i] \subset \mathbb{Q}[i+2^{1/4}]$. Let us call $\alpha = i+2^{1/4}$, then we know 
$$(\alpha-i)^4 - 2 = 0$$
expand the 4th power
$$\alpha^4 - 4\alpha^3 i - 6 \alpha^2 + 4\alpha i + 1 - 2 = 0$$
then we can solve for $i$ in terms of $\alpha$, so $i\in \mathbb{Q}[\alpha]=:\mathbb{Q}[i+2^{1/4}]$. Then $2^{1/4} \in  \mathbb{Q}[i+2^{1/4}]$ as well. 
However the hint is to show the orbit of $i+2^{1/4}$ has more than 5 elements under the action of the galois group $Gal(\mathbb Q[2^{1/4}, i]/\mathbb{Q})$. I calculated the Galois group which is $D_8$, and the orbit has more than 5 elements, but how would conclude from this?
 A: Let $K=\mathbb{Q}[2^{1/4},i]$, $L=\mathbb{Q}[i+2^{1/4}]$, and $G=Gal(\mathbb{Q}[2^{1/4},i]/\mathbb{Q})\cong D_8$.  By Galois theory, in order to show that $L=K$, it suffices to show that the subgroup $H\subseteq G$ of automorphisms that fix $L$ is trivial.  Since $i+2^{1/4}$ generates $L$, we can also describe $H$ as the subgroup of automorphisms that fix $i+2^{1/4}$.  So the orbit of $i+2^{1/4}$ is in bijection with the coset space $G/H$.  
Now if $H$ is nontrivial, it has at least $2$ elements, so $G/H$ has at most $8/2=4$ elements.  So since the orbit of $i+2^{1/4}$ has at least $5$ different elements, $H$ must be trivial, and so $L=K$.
A: To elaborate on @Eric Wofsey 's answer, or perhaps show a more direct view of the injection, define a $G=Gal(K/\mathbb{Q})$-action on $K=\mathbb{Q}[2^{1/4},i]$ by $g.x=g(x)$.
Let us look at the orbit $o(x)$ of $x=2^{1/4}+i$ and the stabilizer $stab_G(x)$.
By orbit-stabilizer theorem $|o(x)|=\frac{|G|}{|stab_G(x)|}$.
Say you prove that $|o(x)|\geq5$ and $|G|=8$. Then necessarily $|stab_G(x)|=1$ and so $H=Aut(K/\mathbb{Q}[x])=\{{e}\}$.
We know by the fundamental theorem of Galois theory that $\mathbb{Q}[x]=Fix(H)=K$, the latter being $K$ since $H$ is trivial.
