Proving $(\alpha + \beta)^{p^n} = \alpha^{p^n} + \beta^{p^n}$ in a finite field 
Prove that if $F$ is a field with $p^n$ elements and $\alpha,\beta \in F$, then
  $$(\alpha + \beta)^{p^n} = \alpha^{p^n} + \beta^{p^n}$$

From Newton identity, we have that
$$(a + b)^n = \sum_{i = 0}^n \binom ni a^i b^{n - i}$$
However, I'm already stuck because in a finite field we cannot necessarily compute the binomial, since we may divide by zero. How do I get around this?
Furthermore, since $F$ has characteristic $p$, isn't it $p^n = 0$? How does it make sense in the first place? I'm a bit confused.
 A: First consider the case $n=1$.  We take the $p$th power as follows:  $(a+b)^p = a^p + {p \choose 1} a^{p-1} b + \cdots + b^p$.  Observe that except for the first and last term in this sum, the coefficient of each term is a multiple of $p$.  Since the field has characteristic $p$, these are zero.  Hence, $(a+b)^p=a^p+b^p$.  Now take the $p$th power again, and repeat this process $n$ times. 
A: It's fine to leave it as an integer
Every ring admits scalar multiplication by integers; you do not need to consider $\binom{n}{i}$ as an element of the finite field.
It's fine to map it into the finite field
We can always uniquely interpret an integer as an element of any ring by applying the unique map $\mathbb{Z} \to R$. If you want to interpret $\binom{n}{i}$ as an element of the ring, that is how you should do so.
Either way, study the coefficient first as an integer
As seen in this answer, there is an explicit formula for how many times $p$ divides $\binom{p^a}{b}$ in the integers:
$$ v_p\left( \binom{p^a}{b} \right) = a - v_p(b) $$
which we can use to show that all of the coefficients except the first and last are divisible by $p$ over the integers.
Then we either argue that we are multiplying by an integer multiple of $p$, or we are multiplying by a finite field element that is $0$.
A: Let $F$ be a finite field with $m$ elements.
By Lagrange's theorem of group theory applied to $F^\times$, we have $z^{m-1}=1$ for all $z \in F$, $z\ne0$.
Therefore, we have $z^m=z$ for all $z \in F$. In particular, $(\alpha + \beta)^{m} = \alpha + \beta = \alpha^{m} + \beta^{m}$.
