A freshman's dream

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 $1 \leq i \leq p-1.$

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: $\left( \sum_{i=1}^n 1 \right)^p = \sum_{r=1}^n (1^p) = \sum_{r=1}^n 1$