Let $K \to L$ be a finite field extension with $[L:K] = n$ and $K$ a finite field of cardinality $q$. A proper subextension is a field extension $K \to M$ such that $M \subsetneq L$. I want to show that $L$ can not be written as the union of its proper subextensions.
I've already shown that I can write any proper subextension $M$ as $M = \{x \in L\;|\; x^{q^m} - x = 0\}$ for some $m \in \mathbb{N}$. Using this, I can show that I can't write $L$ as union of such subextensions because that would lead to a contradiction to the fact that the multiplicative group of a finite field is cyclic (it would imply that $L$ can not contain an element of order $q^n -1$). However, we have not shown that fact in class yet so I'm wondering if I'm overlooking something much simpler.
The subextensions - or intermediate fields - $K \subset M \subset L$ are in bijection with the divisors of $n$, to each $d \mid n$ there corresponds a unique intermediate field $M_d$ with $[M_d : K] = d$ - this is the set of zeros of $X^{q^d} - X$ in $L$. Thus we have
\begin{align} \operatorname{card} \Biggl(\bigcup_{\substack{d\mid n \\ d < n}} M_d\Biggr) &\leqslant \sum_{\substack{d\mid n \\ d < n}} \operatorname{card} M_d \\ &= \sum_{\substack{d\mid n \\ d < n}} q^d\\ &\leqslant \sum_{k = 1}^{n-1} q^k \\ &< q^n, \end{align}
since $q \geqslant 2$.