show that $(n+1)(n+2)...(2n)$ is divisible by $2^n$ but not by $2^{n+1}$ Is this proof correct?
Suppose $2^k$ is the largest power of $2$ in the sequence $n+1, n+2, ... 2n$
Then we can compute the power of 2 in the product as $n/2 + n/2^2 + ... n/2^k = n(1 + 2 + .... 2^{k-1})/2^k = n$. 

We can write as $$(n+1)(n+2)...(2n)=\frac{2n!}{n!}\tag1$$ Now let us  calculate the power of $2$ contained in $2n!$ $$\text{Power of }2 \text{ in } 2!=\frac{2n}{2}+\frac{2n}{2^2}+\frac{2n}{2^3}+\frac{2n}{2^4}\ldots\\=2n\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\ldots\right)\\=2n\left(\frac{\frac{1}{2}}{1-\frac12}\right)\\=2n$$ which means $$2n!=2^{2n}\times a$$ 
Similarly power of $2$ contained in $n!$ $$\text{Power of }2 \text{ in } 2!=\frac{n}{2}+\frac{n}{2^2}+\frac{n}{2^3}+\frac{n}{2^4}\ldots\\=n\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\ldots\right)\\=n\left(\frac{\frac{1}{2}}{1-\frac12}\right)\\=n$$ which means $n!=2^{n}\times b\tag{Note that $b|a$}$  $\therefore$ equation $(1)$ can be written as $$(n+1)(n+2)...(2n)=\frac{2^{2n}\times a}{2^{n}\times b}$$ $$(n+1)(n+2)...(2n)=\frac{2^{n}\times a}{b}$$
