Prove that $\displaystyle 2! \cdot 4! \cdot 6!.....(2n)! \geq \left\[(n+1)!]^n$

I've already proved that P(1) is a true statement.

My inductive hypothesis is that $\displaystyle 2! \cdot 4! \cdot 6!.....(2k)! \geq \left\[(k+1)!]^k$ is true.

I'm trying to show that P(k+1) is true: $\displaystyle 2! \cdot 4! \cdot 6!.....(2k+2)! \geq \left\[(k+2)!]^{k+1}$

$\displaystyle 2! \cdot 4! \cdot 6!.....(2k+2)! $

$\displaystyle \Rightarrow 2! \cdot 4! \cdot 6!.....(2k+2)(2k+1)(2k)! $

$\displaystyle \Rightarrow 2! \cdot 4! \cdot 6!.....(2k)!(2k+2)(2k+1) $

I'm stuck here though. How can this lead to being $\displaystyle \geq \left\[(k+2)!]^{k+1}$?