How do you prove $\lim\_{k\to\infty,k\in\mathbb{N}}{n\choose k}2^{1-{k\choose2}}=1$, where $n=\max\{n\in\mathbb{N}:{n\choose k}2^{1-{k\choose2}}<1\}$? The expression ${n\choose k}2^{1-{k\choose2}}$ comes up in a lower bound for the Ramsey number $R(k,k)$, and I want to fill in the steps in the derivation of the asymptotics of $n$.


I have tried playing around with identities like $${m\choose k+1}2^{1-{k+1\choose2}}=\frac{m-k}{k+1}2^{-k}{m\choose k}2^{1-{k\choose2}}$$
and 
$${m+1\choose k+1}2^{1-{k+1\choose2}}=\frac{m+1}{m-k}{m\choose k+1}2^{1-{k+1\choose2}},$$
but I didn't get anywhere.


*Spencer, Joel*, Ten lectures on the probabilistic method., CBMS-NSF Regional Conference Series in Applied Mathematics. 64. Philadelphia, PA: SIAM, Society for Industrial and Applied Mathematics. vi, 88 p. (1994). [ZBL0822.05060](https://zbmath.org/?q=an:0822.05060).



$\lim_{k\to\infty}{n\choose k}2^{1-{k\choose2}}$, where $n=\max\{n\in\mathbb{N}:{n\choose k}2^{1-{k\choose2}}<1\}$
Let $N(k) = \max \{n \in \mathbb N: \; {n \choose k} 2^{1-{k\choose 2}} < 1 \}$.
Note that $$\frac{{n+1 \choose k}}{{n \choose k}} = \frac{n+1}{n+1-k}$$
so it suffices to prove that $N(k)/k \to \infty$ as $k \to \infty$.
But for any finite $c > 1$, 
$$ \log {c k \choose k} \sim (c \log c - (c-1) \log (c-1)) k = o(k^2)$$
so $$ {c k \choose k} 2^{1 - {k \choose 2}} \to 0 \ \text{as}\ k \to \infty$$
and thus for sufficiently large $k$ we must have $N(k) > c k$.


