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 $$(2n)!$$
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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.
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Hint: You can verify by a computation that $$\frac{1}{n+1}\binom{2n}{n}=\binom{2n}{n}-\binom{2n}{n+1}.$$
Detail: We have $$\frac{1}{n+1}\binom{2n}{n}=\frac{1}{n(n+1)}\frac{(2n)!}{(n-1)!n!}=\left(\frac{1}{n}-\frac{1}{n+1}\right) \frac{(2n)!}{(n-1)!n!} .$$ Now $$\frac{1}{n}\frac{(2n)!}{(n-1)!n!} =\binom{2n}{n}\qquad\text{and}\qquad \frac{1}{n+1}\frac{(2n)!}{(n-1)!n!} =\binom{2n}{n+1} .$$
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Your expression for $(2n)!$ is correct. You can't easily split it down.
You might want to use $$\binom {2n}{n}=\frac {(2n)!}{n!n!}=\frac {n+1}{n}\frac{(2n)!}{(n+1)!(n-1)!}=\frac {n+1}n\binom {2n}{n+1}$$
noting that $n$ and $n+1$ are coprime, or consider $$\binom {2n}{n}-\binom {2n}{n+1}$$
These numbers are called Catalan numbers, and other proofs come from showing that they count discrete objects like the number of different (consistent) ways to insert pairs of brackets in an algebraic expression containing $n$ symbols [eg a(bc); (ab)c for three symbols].
You are correct: $$(2n)! = (2n)(2n-1)(2n-2)\cdots(3)(2)(1)$$
So, for example, if $n= 3$, then $(2n)! = 720$.