<s>[INST] Solution verification of $\frac{1}{(4n^2)!} >\frac{1}{(4(n+1)^2)!}$ inequality \begin{align}
\frac{1}{(4n^2)!} &>\frac{1}{(4(n+1)^2)!}\\
\frac{(4(n+1)^2)!}{(4n^2)!} &>\frac{1}{1}\\
\frac{(4(n^2+2n+1))!}{(4n^2)!} &>1\\
\frac{(4n^2+8n+4)!}{(4n^2)!} &>1\\
\iff \frac{(4n^2+8n+4)}{(4n^2)}&>1\\
4n^2+8n+4&>4n^2\\
8n &> -4\\
n&>-\frac{1}{2}
\end{align}
This implies that the inequality holds for all natural numbers, thus also the primary inequality $$\frac{1}{(4n^2)!} >\frac{1}{(4(n+1)^2)!}$$ holds.
 [/INST] That looks correct, but it is much easier to see that\begin{align}4(n+1)^2>4n^2&\implies(4(n+1)^2)!>(4n^2)!\\&\implies\frac1{(4(n+1)^2)!}<\frac1{(4n^2)!}.\end{align}
</s>