[Community Question] Calculus: How is $\left(\frac{n!}{(2n +1)!!}\right)^2 4^n = \left(\frac{(2n)!!}{(2n +1)!!}\right)^2 ?$

One of our user asked:

How is $$\left(\frac{n!}{(2n +1)!!}\right)^2 4^n = \left(\frac{(2n)!!}{(2n +1)!!}\right)^2 ?$$

Could anyone explain this for me please?