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

You can do it by induction. The base case is easy. For the induction step, suppose the result is true for $n=k$. So we assume that we know that
$$(k+1)(k+2)\cdots(2k)\tag{1}$$
is divisible by $2^k$ but not by $2^{k+1}$. 
Now the product when $n=k+1$ is
$$(k+2)(k+3)\cdots(2k)(2k+1)(2k+2).\tag{2}$$
To get from the product  (1) to the product (2), we multiply (1) by $\frac{(2k+1)(2k+2)}{k+1}=2(2k+1)$. Thus the product (2) has "one more $2$" than the product (1). 
