This is an old result from 2002. See here.

Really nice problem! First, as $n!$ differ from $(n\pm 1)!$ by a factor of at most $2^{O(\log n)}$. Thus it suffices to prove the case $n=2k$. The upper bound is now trivial. Just note that $a \leqslant \tbinom{2k}{k} = \tfrac{(2k)!}{(k!)^2}$. This is clearly less than $\tbinom{2k}{0}+\tbinom{2k}{1}+...+\tbinom{2k}{2k} = 2^n$. Now, we proceed to prove the lower bound. Let $T=\tbinom{2k}{k}$. We prove the following quick lemma. Lemma: $\nu_p(T) \leqslant \log_p(2k)$. Proof: By Kummer's theorem, this is the number of carries when adding $k$ to $k$ in base $p$. As there are at most $\log_p(2k)$ digits, there are at most $\log_p(2k)$ carries, done. Now, any prime $p$ which $\nu_p(T) > 1$ satisfies $p\leqslant\sqrt{2k}$. These primes contribute at most $$\prod_{p\leqslant\sqrt{2k}} p^{\nu_p(T)} \leqslant (2k)^{\sqrt{2k}} = 2^{O(\sqrt{n}\log n)}.$$Dividing $\prod_{p\leqslant\sqrt{2k}} p^{\nu_p(T)}$ out gives the squarefree factor of $\tbinom{2k}{k}$. As $\tbinom{2k}{k} > \frac{2^n}{n+1}$, the remaining factor is $2^{n+o(n)}$ as desired.

MarkBcc168 wrote: First, as $n!$ differ from $(n\pm 1)!$ by a factor of at most $2^{O(\log n)}$. Thus it suffices to prove the case $n = 2k$. Can someone explain this? I looked on the internet for "big o notation" and cant relate to this.

Bump this too

GorgonMathDota wrote: MarkBcc168 wrote: First, as $n!$ differ from $(n\pm 1)!$ by a factor of at most $2^{O(\log n)}$. Thus it suffices to prove the case $n = 2k$. Can someone explain this? I looked on the internet for "big o notation" and cant relate to this. It means you can find a constant c such that $n!-(n-1)!<2^{c.\log n}$ .

ZhangLi wrote: GorgonMathDota wrote: MarkBcc168 wrote: First, as $n!$ differ from $(n\pm 1)!$ by a factor of at most $2^{O(\log n)}$. Thus it suffices to prove the case $n = 2k$. Can someone explain this? I looked on the internet for "big o notation" and cant relate to this. It means you can find a constant c such that $n!-(n-1)!<2^{c.\log n}$ . So, why is it suffices to consider the case for $n = 2k$?

The point is the difference between $(2k)!$ and $(2k\pm 1)!$ are negligable. Therefore it suffices to prove $n=2k$. I could be more explicit if I want but phrasing in this way convey an inituition behind each step.

MarkBcc168 wrote: The point is the difference between $(2k)!$ and $(2k\pm 1)!$ are negligable. Therefore it suffices to prove $n=2k$. I could be more explicit if I want but phrasing in this way convey an inituition behind each step. I see. Thanks for explaining !