TITLE: Show that $ (n+1)(n+2)\cdots(2n)$ is divisible by $ 2^n$, but not by $ 2^{n+1}$
QUESTION [7 upvotes]: darij grinberg's note: 
I. Niven, H. S. Zuckerman, H. L. Montgomery, An introduction to the theory of numbers, 5th edition 1991, §1.4, problem 4 (b)


Let $n$ be a nonnegative integer. Show that $ (n+1)(n+2)\cdots(2n)$ is divisible by $ 2^n$, but not by $ 2^{n+1}$.

I have no idea how to prove this. Can anyone help me through the proof. Thanks.

REPLY [2 votes]: Consider the product to be $P$.
Then $P$ is $\frac{(2n)!}{n!}$.
Now evaluate $(2n)!$ as 
$$(2n)!={(2n)\cdot (2n-2)\cdot \ldots (4)\cdot (2)}{(2n-1)\cdot(2n-3)\ldots 3\cdot 1} .$$
The $(2n) \cdot (2n-2) \cdot ... (4) \cdot (2)$ part of the right hand side can be rewritten as $2^n (n!)$, so we get
$$(2n)!=2^n(n!){(2n-1)\cdot (2n-3)\ldots 3\cdot 1} .$$
Therefore 
$$P = 2^n(\text{product of odd numbers from } 1 \text{ to }(2n-1)) .$$
Hence proved