What is the meaning of $(2n)!$ I came across something that confused me $$(2n)!=?$$ What does this mean: $$2!n!, \quad  2(n!)$$ or $$(2n)!=(2n)(2n-1)(2n-2)...n...(n-1)(n-2)...1$$ 
Which one is right? 
The exercise is to show that $$(n+1)\bigg|\left(\begin{array}{c}2n\\n\end{array}\right)$$Then I thought of using the combination formula $\left(\begin{array}{c}n\\k\end{array}\right)=\frac{n!}{k!(n-k)!}$ to decrease my expression, but then I came across $$(2n)!$$

If you have $(2n)!$, then is is indeed
$$
(2n)! = (2n)(2n-1)(2m-2) \cdots 3\cdot 2\cdot 1.
$$
So, you want to consider
$$
\binom{2n}{n} = \frac{(2n)!}{n!(2n - n)!} = \frac{(2n)!}{(n!)^2} = \frac{(2n)(2n-1)\cdots (n+1)}{n!}
$$
You want to show that 
$$
n+1 \mid \binom{2n}{n}.
$$
That is, you want now to show that $n!\mid (2n)(2n-1)\cdots (n+2)$. You could try to do this using induction. For $n = 1$ (or $n=2$) this is obviously true. Then take it from there.
