Theorem: Let $\mathbb{F}$ be a field and $char(\mathbb{F}) = p$ prime, then $\forall k \in \mathbb{N}$ and $ \forall a, b \in \mathbb{F}: > (a+b)^{p^k} = a^{p^k} + b^{p^k}$
Once the case $k=1$ is proven, then you can prove the full theorem by induction. For $k=1$:
Newton's binomium tells us that $(a+b)^p = \sum_{n=0}^p > \binom{p}{n}a^nb^{p-n}$
If $n \neq 0$ and $n \neq p$ then $\binom{p}{n} = \frac{p(p-1)...(p-n+1)}{n!} \in \mathbb{N}$ which implies $\binom{p}{n}$ is a multiple of $p$ so that $\binom{p}{n}a^nb^{p-n} = 0$.
If $n = 0$ then $\binom{p}{n}a^nb^{p-n} = a^p$
if $n = p$ then $\binom{p}{n}a^nb^{p-n} = b^p$
So that $(a+b)^p = \sum_{n=0}^p \binom{p}{n}a^nb^{p-n} = a^{p} + b^{p}$
I'm not sure about the "If $n \neq 0$ and $n \neq p$ then $\binom{p}{n} = \frac{p(p-1)...(p-n+1)}{n!} \in \mathbb{N}$"
I think it can be shown that the prime factorization of $n!$ is part of the prime factorization of $p(p-1)...(p-n+1)$ but I can not see how.
You don't need the prime factorisation of $n!\;$ to see that. First,as @quid recalled it, binomial coefficients are natural numbers, hence the denominator divides the numerator.
Furthermore $\;n!\;$ is a product of factors, each of which is smaller than $p$. Hence none can divide $p$ since $p$ is prime. Therefore, their product $n!\,$ is coprime with $p$, and divides $p\,\times$ another factor. By Gauß's lemma, it divides this other factor.
Note: the characteristic of a field is always $0$ or a prime number.