Prove that for any $\alpha \in \mathbb{F}_{p^n}$ its trace over $\mathbb{F}_p$ equals $\sum_{i=0}^{n-1} \alpha^{p^i}$.
(You may start from any well known general definition of trace you like, e.g. trace of the matrix corresponding to the map $x\to \alpha x$ or the sum of the Galois conjugates of $\alpha$.)
Let $g$ be a generator of $\mathbb{F}_{p^n}^{*}$. Currently I am aware of the following two observations:
If we prove the claim for each $g^{p^i}$, then the claim follows in general easily by writing $\alpha$ as a linear combination of these and using $(\lambda g^{p^i})^p = \lambda g^{p^{i+1}}$ for $\lambda \in \mathbb{F}_p$.
Here is a simple proof for $\alpha = g$. From the minimal polynomial $m$ of $g$ we get $g^n = \sum_{k=0}^{n-1} a_kg^k$ for some $a_k \in \mathbb{F}_p$, where $a_{n-1}$ is the sum of the roots of $m$ by Vieta. The matrix of the map $x\to gx$ with respect to $(1,g,g^2,\ldots,g^{n-1})$ has zeroes on the diagonal, apart from the last entry which is $a_{n-1}$. Hence the trace equals $a_{n-1}$ and it is $\sum_{i=0}^{n-1} g^{p^i}$ because the map $x \to x^p$ is an automorphism (so $g^{p^i}$ are all roots of $m$ and they are $n = \deg m$ in number, so there can be no more roots).
Any idea if 2) can be generalized to $g^{p^i}$, or maybe another reasonable approach? Any help appreciated!
Starting from the definition of the trace as the sum over all Galois conjugates:
The Galois group of $\Bbb{F}_{p^n}$ over $\Bbb{F}_p$ is cyclic of order $n$, and generated by the Frobenius map $x\ \mapsto\ x^p$. So $$\operatorname{Tr}(\alpha)=\sum_{i=0}^{n-1}\alpha^{p^i}.$$