The following question struck me as pretty interesting: Let $\Bbb F$ be a field of characteristic $p$ (a prime, of course). I'm then asked to show that $|\mathbb{F}| = p^n$ for some $n\geq 1$.
Here's my intuition. Certainly we know that the prime subfield of $\mathbb{F}$ has order $p$. Now if there's an element (treating $\mathbb{F}$ now as a vector space over itself) independent from it, we have the $Span\{1,a_1\}$ as the usual set of linear combinations of $1$ and $a_1$. And any element of a field of characteristic $p$ added to itself $p$ times is $0$, so now we have $p^2$ possible linear combinations. And so on, arguing inductively. Is this argument kosher? Or does more need to be said to make it rigorous?