Binomial Coefficient of Prime/Proof 2

Theorem

Then:

$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$

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

$\dbinom n k \equiv \dbinom {\left \lfloor {n / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor} \dbinom {n \bmod p} {k \bmod p} \pmod p$

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:

$\dbinom {p \bmod p} {k \bmod p} = \dbinom 0 {k \bmod p} = 0$

The result follows immediately.

$\blacksquare$