Binomial Coefficient of Prime/Proof 2

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
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.