$\binom{2n}{n} \equiv 0 \pmod{p}$ 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?
$$\binom{2n}{n}=\frac{(2n)!}{n!n!}=\frac{(n+1)(n+2) \cdots (2n-1)2n}{1 \cdot 2 \cdots n}\overset{ \text{As } 1 \leq n <p<2n }{=} \frac{(n+1)(n+2) \cdots p \cdots (2n-1)2n}{1 \cdot 2 \cdots n} \equiv 0 \pmod p$$

In my opinion, the last line needs a justification, note that the fact that $p$ is a prime is the key here.
Note that when you say 
$$ \frac{(n+1)(n+2) \cdots p \cdots (2n-1)2n}{1 \cdot 2 \cdots n} \equiv 0 \pmod p \,,$$
you say that  $\frac{(n+1)(n+2) \cdots (p-1)(p+1) \cdots (2n-1)2n}{1 \cdot 2 \cdots n}$ is integer, but why is that the case?
The argument is very simple:
$$\binom{2n}{n} n! =(n+1)(n+2) \cdots p \cdots (2n-1)2n$$
Therefore $p$ divides $\binom{2n}{n} n!$. As $p$ is prime, we get $p| \binom{2n}{n}$ or $p|n!$. But $p|n!$ is not possible, as again primality would imply that it divides one of $1,2,3,...,n$.
