The trace of an element $\alpha \in \Bbb F_{q^m}$ over $\Bbb F_q$ is defined to be the sum of all its conjugates, i.e. $$\mathrm{Tr}_{F/K}(\alpha)=\sum_{i=0}^{m-1}\alpha^{q^i}.$$
I have also seen it defined as follows: if $f(x)\in \Bbb F[x]$ of degree $d$ is the minimal polynomial of $\alpha$, then $$g(x)=(f(x))^{\frac{m}{d}}=x^m+a_{m-1}x^{m-1}+\cdots+a_0$$ $$=(x-\alpha)(x-\alpha^q)\cdots(x-\alpha^{q^{m-1}})$$ is its characteristic polynomial and therefore by comparison of the coefficients $\mathrm{Tr}_{F/K}(\alpha)=-a_{m-1}$.
Can anyone clarify why the last equality holds?