Factorial inequality $2!\,4!\,6!\cdots (2n)!\geq\left((n+1)!\right)^n$ using induction

Question

I want to show $2!\,4!\,6!\cdots (2n)!\geq\left((n+1)!\right)^n.$

Assume that it holds for some positive integer $k\geq 1$ and we will prove,

$2!\,4!\,6!\cdots (2k+2)!\geq\left((k+2)!\right)^{k+1}$.

Then we have \begin{eqnarray} 2!\,4!\,6!\cdots (2k+2)! = (2!\,4!\,6!.....(2k)!)((2k+2)^k)((2k+1)^k) \hspace{2cm} (1). \end{eqnarray}

Does (1) hold?

EDIT

So this is what I came up with.

I want to show $2!\,4!\,6!\cdots (2n)!\geq\left((n+1)!\right)^n.$

Assume that it holds for some positive integer $k\geq 1$ and we will prove,

$2!\,4!\,6!\cdots (2k+2)!\geq\left((k+2)!\right)^{k+1}$.

If we represent the LHS as [(2*1)(4*3*2*1)(6*5*4*3*2*1).....((2k+2))((2k+1))((2k!))]

we notice that (2*1)[(4*3)(6*5)......(2k+2)(2k+1)][(2*1)(4*3*2*1).....(2k!)]

Noticing that (2*1)[(4*3)(6*5)......(2k+2)(2k+1)]=(2k+2)! and using the inductive hypothesis we have

[(2*1)(4*3*2*1)(6*5*4*3*2*1).....((2k+2))((2k+1))((2k!))]=> [(2k+2)!] ([(K+1)!]^K)

and then from there we show that it is greater than or equal to [(K+2)!]^(K+1).

Is this correct?

Answer 1

No, $(1)$ does not hold. If $2k+1$ is prime, the left side has one factor of $2k+1$ while the right side has $k$ of them, so they are not equal. I suspect you have not written what you intended, but the editing seems to be correct to the original question.

Answer 2

Notice it holds for $n=1$

You just have to prove $(2n)!\geq \frac{(n+1)!^n}{n!^{n-1}}=(n+1)^{n-1}(n+1)!$

Which amounts to proving:

$2n(2n-1)\dots (n+2)\geq (n+1)^{n-1}$ which we can prove again by induction, by noting it holds for $n=1$, all we have to prove now is.

$2n\geq(n+1)$, which is true.

Answer 3

The claimed inequality holds when $n = 1.$ Suppose that $k >1$ and we know the inequality holds for $k-1.$ Then $2! \ldots (2k-2)! \geq k!^{k-1}.$ To get what we want, it suffices to prove that $(2k)! \geq (k+1)^{k}k!.$ For this factor multiplies the right factor up to $(k+1)!^{k}.$ But the needed inequality holds, because $(2k)!$ is the product of $k!$ with $k$ terms each at least as large as $k+1$ (that is, $k+1,k+2,\ldots,k+(k-1),k+k).$