TITLE: Proof that ${2p\choose p}\equiv 2\pmod p$
QUESTION [9 upvotes]: While trying to prove that ${2p\choose p}\equiv 2\pmod p$, I saw a few successful methods, many of which seemed rather complicated or relied on obscure theorems (at least from the point of view of someone with my limited knowledge). Since I am quite new with number theory, I would appreciate it if someone could tell me if a method I came up with works, or, if not, where it goes wrong. My idea is that by expanding and reducing  ${2p\choose p}$, we get $$2\prod\limits_{i=1}^{p-1} \frac{2p-i}{p-i} \equiv 2\prod\limits_{i=1}^{p-1} 1 \equiv 2 \pmod {p}$$ because $2p-i \equiv p-i \pmod{p}$.
Is this valid? I know division$\pmod{p}$ is slightly weird, especially when the denominator is $\equiv 0\pmod{p}$, so I'm not sure if all of the statements I made are true. Let me know if I need to clarify anything, and thank you in advance for any help you can provide.
P.S. If there are any other [more accurate] proofs of this congruence that aren't too complicated, I'd love to see them.

REPLY [2 votes]: An elementary method to prove this is to note that;
$$p+1\equiv 1\pmod{p}$$
$$p+2\equiv 2\pmod{p}$$
$$\vdots$$
$$2p-1\equiv p-1\pmod{p}.$$
Multiplying all these Equivalences gives us;
\begin{align*}
(p+1)(p+2)\cdots(2p-1)
&\equiv 1\cdot2\cdots(p-1)\pmod{p}\\
&\equiv (p-1)!\pmod{p}.
\end{align*}
Hence, we can say that;
$$p\mid (p+1)(p+2)\cdots(2p-1)-(p-1)!$$
$$\implies p\mid \frac{(p+1)(p+2)\cdots(2p-1)-(p-1)!}{(p-1)!}$$
$$\implies p\mid \frac{(p+1)(p+2)\cdots(2p-1)}{(p-1)!}-1$$
$$\implies \frac{(p+1)(p+2)\cdots(2p-1)}{(p-1)!}\equiv 1\pmod{p}$$
$$\implies \frac{(p+1)(p+2)\cdots(2p-1)(2p)}{(p-1)!(2p)}\equiv 1\pmod{p}$$
$$\implies \frac{(p+1)(p+2)\cdots(2p-1)(2p)}{(p-1)!(p)}\equiv 2\pmod{p}$$
$$\implies \frac{(p+1)(p+2)\cdots(2p-1)(2p)}{(p)!}\equiv 2\pmod{p}$$
$$\implies \frac{(2p)!}{(p)!^2}\equiv 2\pmod{p}$$
$$\implies \binom{2p}{p}\equiv 2\pmod{p}.$$
Hence, we know that $p$ divides $\binom{2p}{p}-2,$ as required.