Proof that $(u+v)^p = u^p + v^p$ in vector space over finite field of characteristic $p$.

Question

Let $\mathbb F_q$ with $q = p^n$ a finite field of characteristic $p$. Then for all $x,y \in \mathbb F_q$ we have $(x+y)^p = x^p + y^p$. If $V$ is a finite-dimensional $\mathbb F_q$-vector space of dimension $n$, then I want to proof that for all $u,v \in V$ we have $$ (u + v)^p = u^p + v^p $$ where $u^p := u + \ldots + u$ ($p$-times) by definition. I know as $V \cong \mathbb F_q^n$ then this holds by "computing in the coordinates", but is there a proof without resorting to coordinates, i.e. computing in $\mathbb F_q^n$?

Answer 1

Because addition is commutative, $$(u+v)^p=\underbrace{(u+v)+\cdots+(u+v)}_p=\underbrace{u+\cdots+u}_p+\underbrace{v+\cdots+v}_p=u^p+v^p$$

Answer 2

The characteristic of the field is $p$, so we have $a^p=0$ for every element $a$ with the given definition. This immediately implies the given equation.

Answer 3

It's true even for vectors $v \in V$ over $\Bbb F_q$: $v^p = \underbrace {v + \dots + v} _ {p \ \text{times}} = (\underbrace {1 + \dots + 1} _ {p \ \text{times}}) \cdot v = 0 \cdot v = 0$.

In fact, this shows not only that $(u+v)^p = u^p + v^p$, but that the equality is trivially true, each term of it being $0$.