Is that right asymptotic formula?

Question

Is it correct asymptotic formula: $ \ln {n^2 \choose n} \sim n\ln{n} $

Answer 1

Hint

$$\log \left(\binom{n^2}{n}\right)=\log\left(\frac{(n^2)!}{n!(n^2-n)!}\right)=\log((n^2)!)-\log(n!)-\log((n^2-n)!)$$ Now, use Stirling approximation $$\log(m!)=m (\log (m)-1)+\frac{1}{2} \left(\log (2 \pi )+\log (m)\right)+O\left(\frac{1}{m}\right)$$ Apply it to each term and simplify.