Binomial Coefficient of Prime

Theorem
Let $$p$$ be a prime number.

Then:
 * $$\forall k \in \Z: 0 < k < p: \binom p k \equiv 0 \left({\bmod\, p}\right)$$

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

Proof
Since $$\binom p k = \frac {p \left({p-1}\right) \left({p-2}\right) \cdots \left({p-k+1}\right)} {k!}$$ is an integer, we have that $$k! \backslash p \left({p-1}\right) \left({p-2}\right) \cdots \left({p-k+1}\right)$$.

But since $$k < p$$ it follows that $$k! \perp p$$, i.e. that $$\gcd \left\{{k!, p}\right\} = 1$$.

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

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

Hence the result.