I have to compute lim.
I tried say that this limit exists and it's l, so we have \lim_{n\rightarrow\infty}\frac{n^n}{(n!)^2} = L then I rewrited it as:
\lim_{n\rightarrow\infty}(\frac{\sqrt n}{\sqrt[n]{n!}})^{2n} then I used natural log over the whole expresion but didn't got into a nice place.
I don't know about Pi function or gamma function so therefore can't really use L'Hospital's rule.
Answer
By ratio test
\frac{(n+1)^{n+1}}{((n+1)!)^2}\frac{(n!)^2}{n^n}=\frac1{n+1}\left(1+\frac1n\right)^n\to 0
then
\lim_{n\rightarrow\infty}\frac{n^n}{(n!)^2}=0
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