Question about the Characteristic of $\mathbb{F}_{p^n}$

Question

We can prove that any finite field of prime characteristic $p$ must have $p^n$ elements.

Conversely, let $F$ be a finite field with $p^n$ elements, where $p$ is a prime number. Is the following argument a correct way to prove that $F$ has characteristic $p$ ?

We can show, without assuming that $F$ has characteristic $p$, that every finite field with $p^n$ elements is the splitting field of $x^{p^n}-x\in\mathbb{F}_p[x]$. Therefore $E/\mathbb{F}_p$ is a field extension and the characteristic of $E$ must equal the characteristic of $\mathbb{F}_p$ which is $p$.

Answer 1

Like @Martín-Blas Pérez Pinilla: said, $$\underbrace{1+ \cdots + 1}_{p^n}=0$$ and so

$$\underbrace{p \times \cdots \times p}_{n}=0$$

in a field, so $p=0$.

Answer 2

The additive group of $\Bbb F_{p^n}$ has order $p^n$, that is...

Answer 3

If $$ \mathbb{F}_q \subseteq \mathbb{F} $$ is a subfield, then $\mathbb{F}$ is a finite-dimensional $\mathbb{F}_q$ vectorspace, so it must have $q^m$ elements. But $p^n=q^m$ implies $p=q$