<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] 
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.
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