how to prove that $\displaystyle \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}= \infty$ without using Stirling's approximation

In my exam there was this question :

Prove convergence or divergence of :

$\displaystyle \sum_{n=1}^ \infty \frac{(n!)^24^n}{(2n)!}$

I noticed that :

$\displaystyle \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}=\frac{4^n 2\pi n \left(\frac{n}{e} \right)^{2n}}{\left(\frac{2n}{e} \right)^{2n }\sqrt{2\pi (2n)}} = \lim_{n \to \infty } \sqrt{\pi n}= \infty$

I couldn't prove this without using Stirling's approximation

I tried to show that $\displaystyle\prod_{k=1}^n \frac{1+\frac{n}{k}}{4} \to 0 $ but that was not possible for me

I also tried to use Stolz cesaro theorem but I couldn't show that the limit $\rightarrow \infty$

(all I got was $\displaystyle \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}= \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!} \times \frac{4(n+1^2)-1}{(2n+1)(2n+2)-1}=\lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}$ )