$\text{Prove That,}\;f(n) = \prod\limits_{i=1}^{n}(4i - 2) = \frac{(2n)!}{n!}$ 
$$\text{Prove That,}\;f(n) = \prod_{i=1}^{n}(4i - 2) = \frac{(2n)!}{n!}$$    

This is a problem from Elementary Number theory.    
My Work:
The statement is true for $i=1$. So, if it is true for $i=k$ it must be true for $ i = k+1$ . I am stuck here, any hint will be helpful.

$$\prod_{i = 1}^n (4i - 2) = \prod_{i = 1}^n 2 \, (2i - 1) = 2^n \, \prod_{i = 1}^n (2i - 1) \frac{\prod\limits_{i = 1}^n 2i}{\prod\limits_{i = 1}^n 2i} = 2^n \frac{\prod\limits_{i = 1}^{2n} i}{2^n \, \prod\limits_{i = 1}^n i} = \frac{(2n)!}{n!}$$
