Right-hand side: ( [1\cdot3\cdot5\cdots(2n-1)(2n+1)]2^n ).
We know [ (2n+1)! = 1\cdot2\cdot3\cdot4\cdots(2n)\cdot(2n+1) = \bigl[1\cdot3\cdot5\cdots(2n+1)\bigr]\cdot\bigl[2\cdot4\cdot6\cdots 2n\bigr] = \bigl[1\cdot3\cdot5\cdots(2n+1)\bigr]\cdot 2^n (1\cdot2\cdot3\cdots n) = \bigl[1\cdot3\cdot5\cdots(2n+1)\bigr]2^n n!. ]
Thus [ \dfrac{(2n+1)!}{n!} = [1\cdot3\cdot5\cdots(2n+1)]2^n. ]
Answer: Proved