annihilator polynomial of a multiplicative group in a Field? Consider the annihilator polynomial of a multiplicative group $H$ of a field $\mathbf{F}_q$.
$$A(x) = \prod_{\alpha\in H} (x-\alpha)$$ 
I read somewhere that this polynomial can be written as $A(x) = x^{|H|}-1$.
How I approached this problem:  Comparing coefficients of $x^0$ we have (disregard sign for now),
$$\prod_{\alpha\in H} \alpha = 1$$
Now $\prod_{\alpha\in H} \alpha = \prod_{\alpha \in H:o(\alpha) \mbox{ divides } 2}\alpha$, since we can cancel every other term with its inverse. Now, the subgroup $K=\{\alpha \in H:o(\alpha) \mbox{ divides } 2\}$ is a multiplicative abelian group and hence must be of the form $K = \mathbf{Z}_2 \times \mathbf{Z}_2  \ldots \times \mathbf{Z}_2 $. But this product does not seem to be zero, intuitively.
For the other coefficients of $x$, the calculation becomes too messy. So, I feel that there must be a slick way of showing this. Can somebody please give me a hint on how to approach this correctly ?
EDIT: clarification $H$ is a subgroup of the multiplicative group and not the multiplicative group of $\mathbf{F}_q$ i.e.  $H \ne \mathbf{F}_q^\star$.
 A: Suppose that $|H|= n$, thus for every $v \in H$ $$\text{ord}(v) \mid n \Longrightarrow v^{n} =1$$
This implies that every element of $H$ is a root of $$p(x) =x^{n}-1 \in F_q[x] $$ But $p(x)$ has degree $n$ and has coefficients in a field; thus it can have at most $n$ roots. 
This means that the roots of $x^{n}-1$ are exactly the elements of $H$ and thus $$\prod_{\alpha\in H} (x-\alpha) = x^{n}-1 = x^{|H|}-1$$ 
A: As $H$ is cyclic it must be  the set of $|H|$th roots of unity...
A: The field $F_q$ is the splitting field of the polynomial $x^q - x$ over the prime subfield $\mathbb{Z}/p\mathbb{Z}$ where $q = p^n$ for some $n \in \mathbb{N}.$ So every $\alpha \in F_q$ is a root of this polynomial. A polynomial of degree $r$ over a field can have at most $r$ roots (counting with multiplicities). Hence the polynomial $x^q - x$ splits into linear factors, i.e. $x^q - x = \prod_{\alpha \in F_q} (x - \alpha).$ Cancelling the factor $x$ (correponding to $\alpha = 0),$ we get $x^{q-1} - 1 = \prod_{\alpha \in F_q^{\times}}(x - \alpha).$
If $H$ is a proper subgroup of $F_q^{\times},$ of order $n$ then consider the polynomial $f(x) = x^n - 1.$ This polynomial can have at most $n$ roots, and all the element of $H$ satisfies this polynomial. So, this polynomial will split into linear factors. Hence $x^n - 1 = \prod_{\alpha \in H}(x - \alpha).$
