<s>[INST] 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?
 [/INST] 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.
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