generator for the mutiplicative group of a field

Proposition 1

The multiplicative group $K^{\mathrm{*}}$ of a finite field $K$ is cyclic.

Theorem 3.1 in the finite fields (http://planetmath.org/FiniteField) entry provesthis proposition along with a more general result:

Proposition 2

If for every natural number $d$, the equation $x^d\mathrm{=}\mathrm{1}$ has at most $d$ solutions in a finite group $G$ then $G$ is cyclic. Equivalently, for any positive divisor $d$ of $\mathrm{|}G\mathrm{|}$.

This last proposition implies that every finite subgroup of the multiplicative group of a field (finite or not) is cyclic.
We will give an alternative constructive proof of Proposition 1:
We first factorize $q-1={\prod }_{i=1}^n{p}_i^{e_i}$. There exists an element $y_i$ in $K^{*}$ such that $y_i$ is not root of $x^{(q-1)/p_i}-1$, since the polynomial has degree less than $q-1$ . Let $z_i=y_i^{(q-1)/p_i^{e_i}}$. We note that $z_i$ has order $p_i^{e_i}$. In fact $z_i^{p_i{}^{e_i}}=1$ and $z_i^{p_i{}^{e_i-1}}=y_i^{(q-1)/p_i}\ne 1$.
Finally we choose the element $z={\prod }_{i=1}^n{z}_i$. By the Theorem 1 here (http://planetmath.org/OrderOfElementsInFiniteGroups), we obtain that the order of $z$ is $q-1$ i.e. $z$ is a generator of the cyclic group $K^{*}$.