Show every $a \in E^*$ is a root of $x^{p^d-1} -1 $? Let $\mathbb{Z}_p < E$ be an extension field of degree $d$. A simple counting argument shows: 
$|E^*| = p^d - 1$ 
Proposition: For all $\alpha \in  E^*$, $x^{p^d-1} -1 = 0.$ 
In a field of $p^d -1 $ elements, does that number equal 0 automatically in a field?
If it does, than $\alpha^0 = 1$ which means 1 - 1 = 0, which means it is a root. If my logic correct?
EDIT:
Okay, so we want to show that every $\alpha \in E^*$ is a zero in $x^{p^d-1} -1$ or $x^{p^d-1} -1 = 0$
How about this:
By Fermat $a^{p-1} \equiv 1\mod p$              (I don't know how to type congruent)
$x^{p^d-1} \equiv 1 \mod p$ 
Where is my understanding lacking so I can deduce from this that any element is 0?

$E^*$ denotes the set of no-zero elements in the field $E$. As $E$ has $p^d$ elements, $E^*$ has $p^d-1$.
Your proposition is false: a power of a non-zero element in a field can't be $0$ since it has an inverse. Are you sure the correct assertion is not $\,x^{p^d-1}=1$, or $\,x^{p^d-1}-1=0\,$?
Edit: This is a sort of generalisation of Fermat, which says that for every element in $\mathbf F_p=\mathbf Z/p\mathbf Z$ satisfies the equation $x^p-x=0$.
Thus, in the field $\mathbf F_{p^2}$, which is the splitting field of any irreducible quadratic polynomial, all elements will satisfy the equation $x^{p^2}-x=0$, and all non-zero elements will satisfy $x^{p^2-1}-1=0$, i. e. they will all be roots of unity.
