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?

Much calculus there :
$$\begin{aligned}\frac{\displaystyle{ \prod_{k =  1}^{2n - 1} k^{\min(k,  2n - k)}}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n -  2k -  1}}} 
&= \frac{\displaystyle{ \prod_{k =  1}^{n} k^{\min(k,  2n - k)}} \times \prod_{k =  n + 1}^{2n - 1} k^{\min(k,  2n - k)}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n -  2k -  1}}} 
\\ &= \frac{\displaystyle{ \prod_{k =  1}^{n} k^k} \times \prod_{k =  n + 1}^{2n - 1} k^{2n - k}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n -  2k -  1}}} 
\\ &= \frac{\displaystyle{ \prod_{k =  1}^{n} k^k} \times \prod_{k =  n + 1}^{2n - 1} k^{2n} \times \prod_{k = 1}^{n - 1}(2k  + 1)^{2k +  1}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n}} \prod_{k =  n + 1}^{2n - 1} k^{k}}
\\ &= \frac{\displaystyle{ \left(\prod_{k =  1}^{n} k^k \right)^2} \times \prod_{k =  n + 1}^{2n - 1} k^{2n} \times \prod_{k = 1}^{n - 1}(2k  + 1)^{2k +  1}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n}} \prod_{k =  1}^{2n - 1} k^{k}}
\\ &= \frac{\displaystyle{ \left(\prod_{k =  1}^{n} k^k \right)^2} \times \prod_{k =  n + 1}^{2n - 1} k^{2n}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k  + 1)^{2n}} \prod_{k =  1}^{n - 1} (2k)^{2k}}
\\ &= \frac{\displaystyle{}\left(\prod_{k =  1}^{2n - 1} k\right)^{2n}}{\displaystyle{}\left(\prod_{k = 1}^{n - 1}(2k  + 1)\right)^{2n}} \frac{\displaystyle{ \left(\prod_{k =  1}^{n} k^k \right)^2}}{\displaystyle{} \prod_{k =  1}^{n} k^{2n} \times \prod_{k =  1}^{n - 1} (2k)^{2k}}
\\ &= \frac{\displaystyle{ \left(\prod_{k =  1}^{n} k^k \right)^2} \times \prod_{k = 1}^{n - 1}(2k)^{2n}}{\displaystyle{}\prod_{k =  1}^{n} k^{2n} \times \prod_{k =  1}^{n - 1} (2k)^{2k}}
\\ &= \frac{\displaystyle{ \prod_{k =  1}^{n} k^{2k}} \times 2^{2n(n-1)}}{\displaystyle{}n^{2n} \times \prod_{k =  1}^{n - 1} (2k)^{2k}}
\\ &= \frac{\displaystyle{ } 2^{2n(n-1)}}{\displaystyle{} \prod_{k =  1}^{n - 1} 2^{2k}}
\\ &= 2^{n(n-1)}
 \end{aligned}$$
