There is a beautiful [paper](http://arxiv.org/abs/1604.08657) on the arXiv by Andrew Suk containing an asymptotic result about the [Erdös-Szekeres convex polygon problem](https://en.wikipedia.org/wiki/Happy_ending_problem). I am struggling with one of the estimates he makes on page 4. He claims that for $n$ *large enough*, we have
$$
2^{n^{4/5} + 4n^{3/5} - 50n^{3/10}} \geq \binom{n + \lceil2n^{3/4} \rceil - 4}{n-2} + 1
$$
Combinatorics is far from being my area of expertise, so I tried proving this the only way I know, which is Stirling's approximation. That did not get me very far. I only obtain that the right hand side grows like $2^n$, which is rather trivial.



> 
> How do you prove that the above inequality holds for large enough $n$? 
> 
> 
> 


PS: I also tried to contact him of course. 



estimating binomial coefficients
We have $$\binom{x}k=\frac{x(x-1)\dots (x-k+1)}{k!}\leqslant \frac{x^k}{k!}\leqslant x^k$$
for positive integers $x, k$. Applying this to $k=[2n^{3/4}]-2$, $x=n+k-2\leqslant 2n$ (for large $n$) we get $$\binom{n+k-2}{n-2}=\binom{n+k-2}{k}\leqslant (2n)^k=e^{k\log(2n)},$$
the rest follows from $4/5>3/4$.


