(This is an extremely useful fact with a straightforward proof. It follows from this, for instance, that any finite extension of a finite field is simple).

Let $\mathbb{K}$ be a finite field, so $|\mathbb{K}| = p^n$ for some prime $p$ and $n \geq 0$. Denote by $\mathbb{K}^\times$ the multiplicative group consisting of the non-zero elements of $\mathbb{K}$. We show that $\mathbb{K}^\times$ is cyclic, i.e., generated by a single element.

Let $d$ be the lowest common multiple (LCM) of the orders of all the elements $a \in \mathbb{K}^\times$. Then $d \mid |\mathbb{K}^\times| = p^n – 1$, and by Lagrange’s theorem, $a^d = a$ for all $a \in \mathbb{K}$, i.e. $$ (x – a) \mid x^{d + 1} – x \quad \forall a \in \mathbb{K}. $$ Hence, $p^n \leq d + 1$, since $x^{d+1} – x$ can have at most $d+1$ roots. Thus $$ p^n – 1 \leq d \leq p^n – 1 $$ i.e. $d = p^n – 1$.

It remains to show that there exists an element whose order is the LCM of the orders of all group elements. Indeed, this is generally true for abelian groups, as shown by the following.

**Proposition:** Let $G$ be an abelian group and let $a, b \in G$ be elements with finite orders $\alpha, \beta$ (respectively). Then there exists $c \in G$ of order $\text{LCM}(\alpha, \beta)$.

**Proof**: First, suppose $\alpha$ and $\beta$ are co-prime and set $c = ab$. Then if $c^i = 1$, then $a^i = b^{-i}$ and so $$ (b^{-1})^{i\alpha} = a^{i\alpha} = (a^\alpha)^i = 1, $$ so $\beta \mid i \alpha$ which implies $\beta \mid i$ by co-primality. Similarly, $\alpha \mid i$. So $\alpha\beta \mid i$ since $\alpha$ and $\beta$ are co-prime, so $c$ has order $\alpha\beta$.

Now consider the general case and write $$ \alpha = \prod_i p_i^{m_i}, \quad \beta = \prod_i p_i^{n_i} $$ where the $p_i$ are distinct primes. For each $i$, define $(M_i, N_i)$ by $$(M_i, N_i) = \begin{cases} (m_i, 0) & \text{if } m_i \geq n_i \\ (0, n_i) & \text{otherwise} \end{cases}$$ and let $$\alpha’ = \prod_i p_i^{M_i},\quad \beta’ = \prod_i p_i^{N_i}.$$ Then $\alpha’ \mid \alpha$ and $\beta’ \mid \beta$, while $\alpha’$ and $\beta’$ are co-prime and $\alpha’ \beta’ = \text{LCM}(\alpha, \beta)$. We have $$ \text{ord}(a^{\alpha / \alpha’}) = \alpha’ \quad \text{and} \quad \text{ord}\left( b^{\beta / \beta’} \right) = \beta’.$$ Thus by the coprime order case first considered, their product $ab$ has order $\alpha’ \beta’ = \text{LCM}(\alpha, \beta)$.