Binomial Coefficient of Prime/Proof 1

Theorem
Let $p$ be a prime number.

Then:
 * $\displaystyle \forall k \in \Z: 0 < k < p: \binom p k \equiv 0 \pmod p$

where $\displaystyle \binom p k$ is defined as a binomial coefficient.

Proof
Since:
 * $\displaystyle \binom p k = \frac {p \left({p-1}\right) \left({p-2}\right) \cdots \left({p-k+1}\right)} {k!}$

is an integer, we have that:
 * $k! \mathop \backslash p \left({p-1}\right) \left({p-2}\right) \cdots \left({p-k+1}\right)$.

But since $k < p$ it follows that:
 * $k! \mathop \perp p$

i.e. that $\gcd \left\{{k!, p}\right\} = 1$.

So by Euclid's Lemma:
 * $k! \mathop \backslash \left({p-1}\right) \left({p-2}\right) \cdots \left({p-k+1}\right)$.

Hence:
 * $\displaystyle \binom p k = p \frac {\left({p-1}\right) \left({p-2}\right) \cdots \left({p-k+1}\right)} {k!}$

Hence the result.