Binomial Coefficient of Prime/Proof 1

Proof
Because:
 * $\dbinom p k = \dfrac {p \left({p - 1}\right) \left({p - 2}\right) \cdots \left({p - k + 1}\right)} {k!}$

is an integer, we have that:
 * $k! \mathrel \backslash p \left({p - 1}\right) \left({p - 2}\right) \cdots \left({p - k + 1}\right)$

But because $k < p$ it follows that:
 * $k! \mathop \perp p$

that is, that:
 * $\gcd \left\{{k!, p}\right\} = 1$

So by Euclid's Lemma:
 * $k! \mathrel \backslash \left({p - 1}\right) \left({p - 2}\right) \cdots \left({p - k + 1}\right)$

Hence:
 * $\dbinom p k = p \dfrac {\left({p - 1}\right) \left({p - 2}\right) \cdots \left({p - k + 1}\right)} {k!}$

Hence the result.