<s>[INST] Proof: $2n! < 2^{2n} (n!)^{2}$ Could someone help with finishing this inequality induction proof.
Hypothesis: $2n! < 2^{2n}(n!)^2$
Step: $2(n+1)! < 2^{2(n+1)}((n+1)!)^2$
(2n+2)! < 2ˆ(2n + 1) * ((n + 1)!)ˆ2
 [/INST] Note that $4n^2+6n+2 < 4n^2+8n+4$ so $\color{blue}{2(n+1)(2n+1)< 2^2(n+1)^2}$.
\begin{eqnarray*}
(2(n+1))!=2(n+1)(2n+1) \underbrace{\color{red}{(2n)!} < \color{blue}{2(n+1)}}_{ \text{by the } {\color{red}{ \text{inductive hypothesis}}}}\color{blue}{(2n+1)} \color{red}{2^{2n}}\underbrace { \color{red}{(n!)^2} < \color{blue}{2^2(n+1)^2}}_{ \text{using the above}} 2^{2n} (n!)^2=2^{2(n+1)} ((n+1)!)^2.
\end{eqnarray*}
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