a) Prove that $\displaystyle \frac{(2n)!}{n!} = 2^n . {1.3.5.7.....(2n-1)}$

b) Prove that $\displaystyle 2.6.10.14......(4n-6)(4n-2)$ equal $\displaystyle (n+1)(n+2)(n+3).....(2n-1)2n$

for part a i have

=$\displaystyle \frac{(2n)(2n-1)(2n-3)......n(n-1)(n-2)(n-3)........}{n(n-1)(n-2)(n-3).....}$

= $\displaystyle (2n)(2n-1)(2n-3).....$

how do i get from $\displaystyle (2n)(2n-1)(2n-3).....$ to

$\displaystyle 2^n . {1.3.5.7.....(2n-1)}$

Please can some tell me how to get to the answer. I am finding this very difficult

for part b i do not know what to do