Binomial Coefficient of Prime

Theorem
Let $p$ be a prime number.

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

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:
 * $k! \backslash p \left({p-1}\right) \left({p-2}\right) \cdots \left({p-k+1}\right)$.

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

Hence the result.

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

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

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

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

by definition of binomial coefficients.

The result follows immediately.