Binomial Coefficient of Prime/Proof 2

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

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

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

Hence, if $0 < k < p$, we have that:
 * $k \bmod p \ne 0$

and so:
 * $\dbinom {p \bmod p} {k \bmod p} = \dbinom 0 {k \bmod p} = 0$

by definition of binomial coefficients.

The result follows immediately.