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

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.
 A: To prove that it is true for $i =n+1$:  
$$\prod_{i=1}^{n+1} (4i-2) = \prod_{i=1}^{n} (4i-2)(4(n+1)-2) = \frac{(2n)!}{n!} (4n+2) =\frac{(2n)!}{n!} 2(2n+1) =\frac{(2n)!}{n!}\frac{2n+2}{n+1}(2n +1) = \frac{(2n+2)!}{(n+1)!} = \frac{(2(n+1))!}{(n+1)!}$$ which was to be proved. Hope it helps.
A: You can prove this without induction also,you just need to note that:
$$\frac {f(n)}{f(n-1)}=4n-2$$ Hence $f(n)=(4n-2)f(n-1)$ Can you complete from here?
A: Assume it is true for $n > 1$. Then we have by induction hypothesis $$\begin{align*}f(n + 1) &= \prod_{i = 1}^{n + 1} (4n - 2)\\ &= (4n + 2)\prod_{i = 1}^{n}(4n - 2)\\ &= (4n + 2)f(n)\\ &= (4n + 2)\frac{(2n)!}{n!}\\ &= 2(2n + 1)\frac{(2n)!}{n!}\\ &= 2(2n + 1)\frac{n+1}{n+1} \frac{(2n)!}{n!}\\ &= \frac{(2n + 2)!}{(n + 1)!}\end{align*}$$
A: $$\prod_{i = 1}^n (4i - 2) = \prod_{i = 1}^n 2 \, (2i - 1) = 2^n \, \prod_{i = 1}^n (2i - 1) \frac{\prod\limits_{i = 1}^n 2i}{\prod\limits_{i = 1}^n 2i} = 2^n \frac{\prod\limits_{i = 1}^{2n} i}{2^n \, \prod\limits_{i = 1}^n i} = \frac{(2n)!}{n!}$$
A: 
Sorry I am not good in latex!!! This way of writing $\frac{2n!}{n!}$ helps in other problems also.
