Consider the polynomials, $f_1, f_2 \in \mathbb{F}_p[x]$, given by
$$ \begin{align} f_1 &= x^{2^n} + 1 \\ f_2 &= x^{2^{n+1}} - 1 \end{align} $$
How can we show that $f_1$ and $f_2$ have isomorphic splitting fields over $\mathbb{F}_p$?
I think I have a bit of a messy way of showing this using the fact that we can write
$$ \begin{align} f_2 &= x^{2^{n+1}} \\ &= (x^{2^n} + 1)(x^{2^{n-1}} + 1) + \cdots + (x^2 + 1)(x + 1)(x - 1) \end{align} $$
However, my idea seems quite inelegant and doesn't use the fact that we are working over a finite field, so I feel like I am missing something.
Any ideas would be greatly appreciated. Thanks in advance!
Indeed, this hasn't much to do with the fact that we're working over a finite field. If you want, consider the question over an arbitrary field $K$. We may suppose $2 \neq 0$, or else there is nothing to prove. Call the corresponding splitting fields $L_1$ and $L_2$. Because $f_1$ divides $f_2$, there is a homomorphism $\phi : L_1 \to L_2$. To show that it is surjective, take a root $\alpha$ of $f_1$ of order $2^{n+1}$. Take any root $\beta$ of $f_2$. Then $\beta$ has order dividing $2^{n+1}$, so it is a power of $\phi(\alpha)$. Hence $\phi$ is surjective.