TITLE: Proof: if $p$ is prime, and $0<k<p$ then $p$ divides $\binom pk$
QUESTION [15 upvotes]: Question : If $p$ is prime, and $0< k< p$ show that $ p \mid {p \choose k}$
${p \choose k}$ can be rewritten as:
$${p(p-1)(p-2)\dots(p-(k-1))(p-k)! \over (p-k)!\cdot k(k-1)(k-2)\dots3\cdot2\cdot1}$$
Now the $(p-k)!$ cancels out. Since in the numerator we have k consecutive  integers,
k divides one of them (not p as p is prime and $k <p$). Then k-1 divides another one or
two terms in the k consecutive terms (again not p). This reasoning continues until 
3 and 2. 
This means that ${p \choose k}$ can be rewritten as pq implying $p | {p \choose k}$
What if any of this proof is wrong. I am worried about any 'overlaps', i.e. say
k-6 divides terms that k-3 divides invalidating the argument.
P.S. I hope the above is legible as I am new to this site. Thank you!

REPLY [2 votes]: Lucas theorem says $\displaystyle \binom n k \equiv \binom {\left \lfloor {n / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor} \binom {n \bmod p} {k \bmod p} \pmod p$, let $p=n$ then we have $$\displaystyle \binom p k \equiv \binom {\left \lfloor {p / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor} \binom {p \bmod p} {k \bmod p} \pmod p$$ and $p\equiv0\pmod p$ , if $0 < k < p$ we have $k \bmod p \ne 0$ but $$\displaystyle \binom {p \bmod p} {k \bmod p} = \binom 0 {k \bmod p} = 0$$ from which the result is derived