Stirling, asymptote of $n^2+n-1\choose n$

Question

I need to find the asymptote of $n^2+n-1\choose n$. any idea? I used the Stirling formula for $n!$ but I found an unexpected final answer.

Answer 1

Let's try

Using $\log(n!) = n \log(n) -n + O(\log(n))$ I get: $$ \log {n^2+n-1\choose n} = (n^2 +n-1) \log(n^2+n-1) - (n^2 +n-1) - n \log(n) + n \\- (n^2-1)\log(n^2-1) + n^2 -1+ O(\log(n^2))=\\= (n^2 -1) \log(\frac{n^2 +n-1}{n^2 -1}) + n\log(\frac{n^2 +n-1}{n}) + O(\log(n^2))$$

But

$$\log(\frac{n^2 +n-1}{n^2 -1})= \frac{1}{n} + O(n^{-2})$$

$$\log(\frac{n^2 +n-1}{n})=\log(n)+O(1)$$

So $$ \log {n^2+n-1\choose n} \approx n + n \log(n)$$