Show that $\frac{(2n)!}{(n)!}=2^n(2n-1)!!$

Question

Show that $\frac{(2n)!}{(n)!}=2^n(2n-1)!!$ is the question I am struggling with.

I started by saying: $(2n)!=2n(2n-1)(2n-2)(2n-3)...3*2*1$

But then I'm stuck.

Answer 1

$$(2n)!=\prod_{k=1}^{2n}k=\prod_{k=1}^{n}(2k)\cdot\prod_{k=1}^{n}(2k-1)=2^n\cdot\prod_{k=1}^{n}k\cdot\prod_{k=1}^{n}(2k-1)=2^n\cdot n!\cdot(2n-1)!!$$

Answer 2

$$\frac{2n!}{n!} = \frac{1\cdot2\cdot\ldots\cdot2n}{1\cdot2\cdot\ldots \cdot n}=\frac{1\cdot3\cdot\ldots\cdot(2n-1)\cdot2\cdot4\cdot\ldots\cdot2n}{1\cdot2\cdot\ldots \cdot n} = \frac{1\cdot3\cdot\ldots(2n-1)\cdot2^n\cdot 1\cdot2\cdot\ldots \cdot n}{1\cdot2\cdot\ldots \cdot n} = 2^n\cdot1\cdot3\cdot\ldots \cdot (2n-1)=2^n(2n-1)!!$$