Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$ Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$
For $n=1$ inequality holds.
$(*)$For $n=k$
$2!\cdot\cdot\cdot (2k)!\ge ((k+1)!)^k$
Multiplying LHS and RHS with $(2k+2)!$ gives
$$2!\cdot\cdot\cdot (2k)!(2k+2)!\ge ((k+1)!)^k(2k+2)!$$
Assume (by contradiction)$$2!\cdot\cdot\cdot (2k)!(2k+2)!< ((k+1)!)^k(2k+2)!$$
$$2!\cdot\cdot\cdot (2k)!(2k+2)!-((k+1)!)^k(2k+2)!<0$$
$$(2k+2)!(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)<0$$
$(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)\ge 0$ by $(*)$, thus inequality holds $\forall n\in\mathbb{N}$
Is this proof correct?
 A: First, show that this is true for $n=1$:
$\prod\limits_{k=1}^{1}(2k)!\geq(1+1)!^{1}$
Second, assume that this is true for $n$:
$\prod\limits_{k=1}^{n}(2k)!\geq(n+1)!^{n}$
Third, prove that this is true for $n+1$:
$\prod\limits_{k=1}^{n+1}(2k)!=$
$\color\red{\prod\limits_{k=1}^{n}(2k)!}\cdot(2(n+1))!\geq$
$\color\red{(n+1)!^{n}}\cdot(2(n+1))!=$
$(n+1)!^{n}\cdot(2n+2)!=$
$(n+1)!^{n}\cdot(n+1)!\cdot\dfrac{(2n+2)!}{(n+1)!}=$
$(n+1)!^{n+1}\cdot\dfrac{(2n+2)!}{(n+1)!}=$
$(n+1)!^{n+1}\cdot\underbrace{(n+2)\cdot(n+3)\cdot\ldots\cdot(2n+2)}_{n+1\text{ times}}\geq$
$(n+1)!^{n+1}\cdot\underbrace{(n+2)\cdot(n+2)\cdot\ldots\cdot(n+2)}_{n+1\text{ times}}=$
$(n+1)!^{n+1}\cdot(n+2)^{n+1}=$
$(n+2)!^{n+1}$

Please note that the assumption is used only in the part marked red.
A: 
Assume (by contradiction)
  $$2!⋅⋅⋅(2k)!(2k+2)!<((k+1)!)k(2k+2)!$$

Why would you assume that, you just showed the contrary one line above ?
Once you're here
$$2!\cdot\cdot\cdot (2k)!(2k+2)!\ge ((k+1)!)^k(2k+2)!$$
you just have to show that $((k+1)!)^k(2k+2)! \ge (k+2)!^{k+1}$.
$(k+2)!^{k+1} = (k+1)!^k \times (k+2)^k \times (k+2)!$ and $(k+2)^k \times (k+2)! \le (2k+2)!$ (because $k+2 < k+2+i$ for all $i$). Thus, you have your result.
A: Define $S$ the wanted number. Then
$$
S=\prod_{i=1}^n (2i)!=\prod_{i=1}^n (2(n+1-i))!.
$$
Therefore, based on $a!b!\ge \max(a,b)!\ge \frac{a+b}{2}!$ whenever $a,b$ are integers with same parity, we conclude that
$$
S=\sqrt{\prod_{i=1}^n(2(n+1-i))!(2i)!}\ge \sqrt{\prod_{i=1}^n(n+1)!^2}=(n+1)!^n.
$$
A: That is equivalent to:
$$ \sum_{k=1}^{n} \log\Gamma(2k+1) \geq n \log\Gamma(n+2) \tag{1}$$
that is a trivial inequality, since $\log\Gamma$ is a convex function, due to the Bohr-Mollerup theorem, for instance.
A: Also, because for odd $n$ we have
$$\frac{(2n)!}{(n+1)!(n-1)!}\cdot\frac{(2n-2)!}{(n+1)!(n-3)!}\cdot...\cdot\frac{(n+1)!}{(n+1)!0!}\cdot(n-1)!(n-3)!...2!\geq1.$$
For the even $n$ it's the similar.
