A freshman's dream If $p$ is prime, then $(x+y)^p=x^p+y^p$ holds in any field of characteristic $p$. However all the proofs I have seen use induction and some relatively nasty algebra despite how fundamental this fact seems.
What is the nicest, "highest level proof" you know?
 A: The binomial coefficient $\binom p  i$ is divisible by $p$ for $1 \leq i \leq p-1$
One way of seeing this is Legendre's formula on the power of a prime dividing some factorial, http://www.cut-the-knot.org/blue/LegendresTheorem.shtml
and
http://en.wikipedia.org/wiki/Factorial#Number_theory
From the formula, $p$  divides $p!$ with exponent exactly $1,$ but $p$ does not divide $i!$ or $(p-i)!$ when $1 \leq i \leq p-1.$
A: Let $F$ be a field of characteristic $p$. Let $f = (1 + x)^p \in F[x]$. We want to show that $f = 1 + x^p$. 
Take the formal derivative: $f' = p(x+1)^{p-1} = 0$
Now we know that $f$ has degree $p$, and its derivative is $0$, so $f$ must be in the form $A + Bx^p$ with $A$, $B \in F$.
$f(0) = 1$ so $A = 1$.
A product of monic polynomials is always monic so $B = 1$.
Q.E.D.
The "freshman's dream" is a corollary of this fact.
The fact that the binomial coefficient $\binom p  i$ is divisible by $p$ for $1 \leq i \leq p-1$ is also a corollary.
The binomial theorem itself can be proved by taking derivatives of $(1 + x)^n$. 
Fermat's little theorem follows easily: $\left( \sum_{i=1}^n 1 \right)^p = \sum_{r=1}^n (1^p) = \sum_{r=1}^n 1$
