Lost on Galois Group over $\mathbb{Q}(i)$ Question. The goal is to find the Galois Group of $t^8-i$ over $\mathbb{Q}(i)$ where $i=\sqrt{-1}$.
I've found the roots of the polynomial because I know I'll need them and they are 
$\pm i^\frac{1}{8}$,$\pm i^\frac{5}{8}$$\pm i^\frac{9}{8}$$\pm i^\frac{13}{8}$.
I need to show that the given polynomial $t^8-i$ is irreducible over $\mathbb{Q}(i)$ right? Is it enough to show that no product of any two roots and no single root is in $\mathbb{Q}(i)$ like I would for a polynomial over $\mathbb{Q}$?
Since I have a root that looks like $i^\frac{1}{8}$ am I right in thinking that the splitting field is then $\mathbb{Q}(i^\frac{1}{8},i)$ and thus the extension is order 8 over $\mathbb{Q}(i)$?.
Also the elements of the Galois Group fix $\mathbb{Q}(i)$ and send a root of $t^8-i$ to another root of $t^8-i$ it seems kind of easy to see that there is something that can send $i^\frac{1}{8} \to -i^\frac{1}{8}$ but how could I find if there is something that performs a more complicated swap like $i^\frac{1}{8} \to -i^\frac{13}{8}$ is brute force verifying that the has the correct properties the only way?
If someone can point me in the right direction it would be most appreciated.
Update: Am I right in thinking that the Galois Group is isomorphic to the Quaternion Group? I think I've identified 3 elements that seem generate different order 4 subgroups and an element that generates an order 2 subgroup which seems to rule out all other groups of order 8.

Well, an eighth root of $i$ is a $32$-nd root of unity, right? And I think you can see that by adjoining one such root of $i$, you get all $32$-nd roots of $1$. So you should think about the field of these roots of $1$, both over $\mathbb Q$ and over $\mathbb Q(i)$. This is conceptually much simpler, I’m sure you’ll agree.
And let me add: if you want to do explicit computations, you might want to work with $\zeta=\exp(\pi i/16)$, clearly a primitive $32$-nd root of unity, and its powers.
