Prove that given fraction is power of two I do not really know where to start with the following:

Prove that
  $$\frac{\displaystyle{ \prod_{k =  1}^{2n - 1} k^{\min(k,  2n - k)}}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n -  2k -  1}}}$$ is a power of two.

Could you give me a hint which helps to solve this problem?
 A: Much calculus there :
$$\begin{aligned}\frac{\displaystyle{ \prod_{k =  1}^{2n - 1} k^{\min(k,  2n - k)}}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n -  2k -  1}}} 
&= \frac{\displaystyle{ \prod_{k =  1}^{n} k^{\min(k,  2n - k)}} \times \prod_{k =  n + 1}^{2n - 1} k^{\min(k,  2n - k)}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n -  2k -  1}}} 
\\ &= \frac{\displaystyle{ \prod_{k =  1}^{n} k^k} \times \prod_{k =  n + 1}^{2n - 1} k^{2n - k}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n -  2k -  1}}} 
\\ &= \frac{\displaystyle{ \prod_{k =  1}^{n} k^k} \times \prod_{k =  n + 1}^{2n - 1} k^{2n} \times \prod_{k = 1}^{n - 1}(2k  + 1)^{2k +  1}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n}} \prod_{k =  n + 1}^{2n - 1} k^{k}}
\\ &= \frac{\displaystyle{ \left(\prod_{k =  1}^{n} k^k \right)^2} \times \prod_{k =  n + 1}^{2n - 1} k^{2n} \times \prod_{k = 1}^{n - 1}(2k  + 1)^{2k +  1}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n}} \prod_{k =  1}^{2n - 1} k^{k}}
\\ &= \frac{\displaystyle{ \left(\prod_{k =  1}^{n} k^k \right)^2} \times \prod_{k =  n + 1}^{2n - 1} k^{2n}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n}} \prod_{k =  1}^{n - 1} (2k)^{2k}}
\\ &= \frac{\displaystyle{}\left(\prod_{k =  1}^{2n - 1} k\right)^{2n}}{\displaystyle{}\left(\prod_{k = 1}^{n - 1}(2k  + 1)\right)^{2n}} \frac{\displaystyle{ \left(\prod_{k =  1}^{n} k^k \right)^2}}{\displaystyle{} \prod_{k =  1}^{n} k^{2n} \times \prod_{k =  1}^{n - 1} (2k)^{2k}}
\\ &= \frac{\displaystyle{ \left(\prod_{k =  1}^{n} k^k \right)^2} \times \prod_{k = 1}^{n - 1}(2k)^{2n}}{\displaystyle{}\prod_{k =  1}^{n} k^{2n} \times \prod_{k =  1}^{n - 1} (2k)^{2k}}
\\ &= \frac{\displaystyle{ \prod_{k =  1}^{n} k^{2k}} \times 2^{2n(n-1)}}{\displaystyle{}n^{2n} \times \prod_{k =  1}^{n - 1} (2k)^{2k}}
\\ &= \frac{\displaystyle{ } 2^{2n(n-1)}}{\displaystyle{} \prod_{k =  1}^{n - 1} 2^{2k}}
\\ &= 2^{n(n-1)}
 \end{aligned}$$
A: Let $f(n)$ be the given expression. Then $f(1) = 1$ and $f(n+1)/f(n) = 2^{2n}$:
$$\frac{f(n+1)}{f(n)} = \frac{(n+1)^{n+1}(n+2)^n \dotsb (2n-1)^3(2n)^2(2n+1)}{3^{2n-1} \dotsb (2n-1)^3(2n+1)} \cdot \frac{3^{2n-3} \dotsb (2n-1)}{(n+1)^{n-1}(n+2)^{n-2} \dotsb (2n-1)}$$
which simplifies to 
$$\frac{f(n+1)}{f(n)} = \frac{(n+1)^2(n+2)^2 \dotsb (2n-1)^2(2n)^2}{3^2 \dotsb (2n-1)^2} = \frac{(n+1)^2(n+2)^2 \dotsb (2n-1)^2(2n)^2 2^2 4^26^2\dotsb (2n)^2} {3^2 \dotsb (2n-1)^2 2^2 4^26^2\dotsb (2n)^2} = \frac{2^{2n} (2n)!^2}{(2n)!^2} = 2^{2n}$$
Then $$f(n) = f(1)\cdot \frac{f(2)}{f(1)} \dotsb \frac{f(n)}{f(n-1)} = 2^{2\cdot 0}  \cdot 2^{2\cdot 1} \dotsb 2^{2\cdot (n-1)} = 2^{2(1+2+\dotsb (n-1))} = 2^{n(n-1)}.$$
