Let $F$ a finite field of characteristic $p$. Show that $p-1$ divides $|F|-1$. (We shall see later that $|F|$ is a power of $p$.)
I am able to solve this by first showing $|F|$ is a power of $p$. If $q$ divides $|F|$ for another prime $q$, then by Cauchy's theorem some element $x$ has order $q$ in the additive group, so $qx=0$. But $qx=px=0$, so the order of $x$ has order $\gcd(p,q)=1$, a contradiction. Thus $|F|=p^n$ for some $n$, and $p-1$ divides $|F|-1$ since $p^n-1=(p-1)(p^{n-1}+\cdots+1)$.
The remark at the end of the exercise that we shall later see $|F|$ is a prime power suggests to me that there may be an alternative proof that doesn't resort to this fact. Is there some obvious observation I'm overlooking, because the proof I believe I have seems simple enough.