Proof that ${2p\choose p}\equiv 2\pmod p$ 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.
 A: The easy way: in characteristic $p$, we have
$$(x+y)^{2p} = (x^p+y^p)^2= x^{2p} + 2x^py^p + y^{2p}.$$
Thus ${2p \choose p} \equiv 2 \mod p$.
A: Incidentally, I just heard about this today. More can be said, in fact $p^2\mid \binom{2p}p-2$. Consider the set $A=\{x,y\}\times \Bbb Z_p$, and the collection $\mathcal C$ of subsets of $A$ of cardinality $p$, which you can think about boards of size $2\times p$ with entries in $\Bbb Z_p$ such that exactly $p$ entries are checked and $p$ are not, where we check the entry $(x,n)$ if $(x,n)\in S\subseteq A$. 
We can make $\Bbb Z_p\times \Bbb Z_p$ act on the set $A$ as follows: if $(a,b)\in\Bbb Z$, we cyclically  permute the first row by $a$ places and the second row by $b$ places. Concretely, if $S$ is a collection of $p$ tuples of the form $(x,n),(y,m)$ then $(a,b)S=$ is the collection of tuples of the form $(x,n+a),(y,m+b)$. One can see that if $C\in \mathcal C$ is not the board obtained by checking all in the first row and nothing in the last, or checking all in the second row and nothing in the last, ${\rm stab}\, a=1$. Thus, if we remove these two elements from $\mathcal C$, which has cardinality $\binom{2p}p$, we get a set of size $\binom {2p}p-2$ which by the orbit-stabilizer theorem is partitioned into sets of size $p^2=|\Bbb Z_p\times \Bbb Z_p|$, which proves the claim.
A: Another "easy way" by Lucas' theorem:
$${2p \choose p} \equiv {2 \choose 1} \equiv 2 \pmod p$$
I'll elaborate since you indicate that you are new to number theory. Lucas' theorem says the following: Let $m$ and $n$ be numbers expressed in base $p$, and let $m_i$ and $n_i$ correspond to the digit in the $i$th place of $m$ and $n$, respectively. Then ${m \choose n} = \displaystyle \prod_i^n {m_i \choose n_i} \pmod p$
In this case, in base $p$, $2p = 20_p$ and $p = 10_p$. Therefore,
$${2p \choose p} \equiv {2 \choose 1}{0 \choose 0} \equiv 2 \pmod p$$
A: 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.
