$\binom{2n}{n} \equiv 0 \pmod{p}$ Let a prime $p$ such that $1 \leq n <p<2n$. Prove that:
$$\binom{2n}{n} \equiv 0 \pmod{p}$$
Can I do it like that?
$$\binom{2n}{n}=\frac{(2n)!}{n!n!}=\frac{(n+1)(n+2) \cdots (2n-1)2n}{1 \cdot 2 \cdots n}\overset{ \text{As } 1 \leq n <p<2n }{=} \frac{(n+1)(n+2) \cdots p \cdots (2n-1)2n}{1 \cdot 2 \cdots n} \equiv 0 \pmod p$$

Lucas' theorem solved this immediately:
$$\binom{n}{i} \equiv \binom{2n-p}{n}\binom{1}{0} \equiv 0 \pmod{p},$$
because $2n-p<n$.
