I want a proof verification.
Prove that $$\boxed{ n\in \mathbb{Z}^{+} \Rightarrow (2n)!\lt 2^{2n}(n!)^{2}}$$ by mathematical induction.
Since $n=1$ implies that $(2\cdot 1)!=2\lt 2^{2\cdot 1}\cdot (1!)^2=2^{2}\cdot 1=4$ the base case holds.
Assume that $(2n)!\lt 2^{2n}(n!)^{2}$ is true for $n$. We now wish to show that this inequality is also true for $n+1$, namely that $$(2(n+1))!\lt 2^{2(n+1)}((n+1)!)^{2}.$$
Multiply the inequality $(2n)!\lt 2^{2n}(n!)^{2}$ by $(2(n+1))^{2}$ and we get that $$(2n)!\cdot (2(n+1))^{2}\lt 2^{2n}(n!)^{2}\cdot (2(n+1))^{2} \Leftrightarrow \\ (2n)!\cdot (2(n+1))^{2}=(2(n+1))!\cdot 2(n+1)\lt 2^{2n}(n!)^{2}\cdot (2(n+1))^{2}=2^{2n}\cdot 2^{2}\cdot (n!\cdot (n+1))^{2}=2^{2(n+1)}((n+1)!)^{2}.$$ That is $$(2(n+1))!\cdot 2(n+1)\lt 2^{2(n+1)}((n+1)!)^{2}.$$ Since $2(n+1)\gt 0$, $\forall n\geq 0$ we have that $$(2(n+1))!\lt (2(n+1))!\cdot 2(n+1)\lt 2^{2(n+1)}((n+1)!)^{2},$$ so $$(2(n+1))!\lt 2^{2(n+1)}((n+1)!)^{2}$$ which we wanted to show.
