# Order of finite fields is pnp^n

Let $$FF$$ be a finite field.
How do I prove that the order of $$FF$$ is always of order $$pnp^n$$ where $$pp$$ is prime?

Let $$pp$$ be the characteristic of a finite field $$FF$$.$$Note 1{}^{\text{Note 1}}$$ Then since $$11$$ has order $$pp$$ in $$(F,+)(F,+)$$, we know that $$pp$$ divides $$|F||F|$$. Now let $$q≠pq\neq p$$ be any other prime dividing $$|F||F|$$. Then by Cauchy’s Theorem, there is an element $$x∈Fx\in F$$ whose order in $$(F,+)(F,+)$$ is $$qq$$.

Then $$q⋅x=0q\cdot x=0$$. But we also have $$p⋅x=0p\cdot x=0$$. Now since $$pp$$ and $$qq$$ are relatively prime, we can find integers $$aa$$ and $$bb$$ such that $$ap+bq=1ap+bq=1$$.

Thus $$(ap+bq)⋅x=x(ap+bq)\cdot x=x$$. But $$(ap+bq)⋅x=a⋅(p⋅x)+b⋅(q⋅x)=0(ap+bq)\cdot x=a\cdot(p\cdot x)+b\cdot(q\cdot x)=0$$, giving $$x=0x=0$$, which is not possible since $$xx$$ has order at least $$22$$ in $$(F,+)(F,+)$$.

So there is no prime other than $$pp$$ which divides $$|F||F|$$.

Note 1: Every finite field has a characteristic $$p∈Np\in\mathbb N$$ since, by the pigeonhole principle, there must exist distinct $$n1 both in the set $${1,2,…,|F|+1}\{1, 2, \dots, \lvert F\rvert +1\}$$ such that $$1+1+⋯+1⏟n1=1+1+⋯+1⏟n2,\underbrace{1+1+\dots+1}_{n_1}=\underbrace{1+1+\dots+1}_{n_2},$$ so that $$1+1+⋯+1⏟n2−n1=0\underbrace{1+1+\dots+1}_{n_2-n_1}=0$$. In fact, this argument also implies $$p≤np\le n$$.