Let $L=\mathbb{F}_{p^n}$ be a finite field with $p^n$ elements, where $p$ is a prime number. We can regard this as a field extension of $K=\mathbb{F}_p$ (a finite field with $p$ elements) of degree $n$. If we consider the field trace $Tr_{L/K}(\alpha)$ for some $\alpha \in L$, by definition his is known to be the trace of the matrix $M_\alpha$ such that $M_\alpha: b \mapsto \alpha b$
As this is a Galois extension, we can alternatively define this as:
$$1.) \hspace{0.5in} Tr_{L/K}(\alpha)=\sum_{\sigma\in Gal(L/K)} \sigma(\alpha)$$
Then each $\sigma(\alpha)$ is simply a conjugate (or just another root) of the irreducible polynomial of $\alpha$ over $K$.
It is known that for all $\alpha \in L$, we have the relationship:
$$2.) \hspace{0.5in} \alpha^{\varphi(p^n)} = \alpha^{p^{n-1}(p-1)} = 1$$
Where $\varphi$ is Euler's totient function. I have also seen equation 1.) also written as :
$$3.) \hspace{0.5in} Tr_{L/K}(\alpha)=\sum_{i=0}^{n-1} \alpha^{p^i}$$
My question is then this: how does one use equation 2.) to construct a polynomial (which is then the irreducible monic polynomial for $\alpha$) for which equations 1.) and 3.) agree?
EDIT: I have seen some proofs where the authors start with the equation $X^{p^{n-1}}+...+X^p+X$, and then they show that this is an $K$-linear, surjective map from $L$ to $K$, as is required for the trace. Is this sufficient to prove that this is the trace?