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.

Alternative Proof
Lucas' Theorem gives:
 * $$\binom n k \equiv \binom {\left \lfloor {n / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor} \binom {n \, \bmod \, p} {k \, \bmod \, p} \pmod p$$

So, substituting $$p$$ for $$n$$:
 * $$\binom p k \equiv \binom {\left \lfloor {p / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor} \binom {p \, \bmod \, p} {k \, \bmod \, p} \pmod p$$

But $$p \, \bmod \, p = 0$$ by definition.

Hence, if $$0 < k < p$$, we have that $$k \, \bmod \, p \ne 0$$, and so:
 * $$\binom {p \, \bmod \, p} {k \, \bmod \, p} = \binom {0} {k \, \bmod \, p} = 0$$

by definition of binomial coefficients.

The result follows immediately.