TITLE: Bounding ${n \choose k}$
QUESTION [8 upvotes]: Let $k > 0$, and $n > 2k$.  Why is it necessarily true that
$${n \choose k} > \frac{n^k}{2^k k!}$$
And is the condition $n > 2k$ necessary for this inequality to hold?

REPLY [6 votes]: If $k\le\frac12n$, then $n-k+1>\frac12n$. Therefore,
$$
\begin{align}
\binom{n}{k}
&=\frac{n(n-1)(n-2)\dots(n-k+1)}{k!}\\
&>\frac{(n/2)^k}{k!}\\
&=\frac{n^k}{2^kk!}\tag{1}
\end{align}
$$
However, the condition that $k<\frac12n$ can be extended to $k\le\frac34n$ by noting that for $0\le j\le k-1$, $(n-j)\color{red}{(n+j-k+1)}\ge n\color{red}{(n-k+1)}>n^2/4$.
$$
\begin{align}
\binom{n}{k}^2
&=\frac{n\color{red}{(n-k+1)}(n-1)\color{red}{(n-k+2)}(n-2)\dots(n-k+1)\color{red}{n}}{k!^2}\\
&>\frac{(n^2/4)^k}{k!^2}\tag{2}
\end{align}
$$
Taking the square root of $(2)$ yields
$$
\binom{n}{k}>\frac{n^k}{2^kk!}\tag{3}
$$
Take It To The Limit:
Above, it is shown that the condition $k<\frac12n$ in the question can be extended to $k\le\frac34n$. One might ask how far this might be pushed. That is, what is the largest $\alpha$ so that $k<\alpha n$ implies $(3)$.
Multiplying $(3)$ by $k!$ and taking the $\log$ of both sides yields
$$
\sum_{j=0}^{k-1}\log(n-j)>k\log(n)-k\log(2)\tag{4}
$$
Noting that
$$
\sum_{j=0}^{k-1}\log(n-j)\ge\int_{n-k}^n\log(x)\,\mathrm{d}x\tag{5}
$$
We get that
$$
\int_{n-k}^n\log(x)\,\mathrm{d}x>k\log(n)-k\log(2)\tag{7}
$$
implies $(3)$. Computing the integral in $(7)$ yields
$$
n\log(n)-n-(n-k)\log(n-k)+(n-k)>k\log(n)-k\log(2)\tag{8}
$$
Subtracting $k\log(n)-k$ from both sides and dividing by $k$ yields
$$
\frac{n-k}{k}\log\left(\frac{n}{n-k}\right)>1-\log(2)\tag{9}
$$
Substituting $\alpha=k/n$ yields
$$
\frac{1-\alpha}{\alpha}\log\left(\frac{1}{1-\alpha}\right)>1-\log(2)\tag{10}
$$
Here is a plot of $\frac{1-\alpha}{\alpha}\log\left(\frac{1}{1-\alpha}\right)$ and $1-\log(2)$:
$\hspace{4cm}$
We can solve $\frac{1-\alpha}{\alpha}\log\left(\frac{1}{1-\alpha}\right)=\beta$ using the Lambert W function:
$$
\begin{align}
\frac{1-\alpha}{\alpha}\log\left(\frac{1}{1-\alpha}\right)&=\beta\tag{11}\\
(1-\alpha)^{1-\alpha}e^\beta&=e^{(1-\alpha)\beta}\tag{12}\\
(1-\alpha)e^{\beta/(1-\alpha)}&=e^\beta\tag{13}\\
-\beta/(1-\alpha)e^{-\beta/(1-\alpha)}&=-\beta e^{-\beta}\tag{14}\\
-\frac{\beta}{1-\alpha}&=W\left(-\beta e^{-\beta}\right)\tag{15}\\
\alpha&=1+\frac{\beta}{W\left(-\beta e^{-\beta}\right)}\tag{16}
\end{align}
$$
$(12)$: multiply by $-\alpha$, exponentiate, multiply by $e^\beta$
$(13)$: raise to the $1/(1-\alpha)$ power
$(14)$: take reciprocals, multiply by $-\beta$
$(15)$: apply $W$
$(16)$: algebraically solve for $\alpha$
Plugging $\beta=1-\log(2)$ into $(16)$ gives
$$
\begin{align}
\alpha
&=1+\frac{1-\log(2)}{W\left(-\frac2e(1-\log(2))\right)}\\
&\approx 0.868708579475419252
\end{align}
$$
Thus, if $k<\alpha n$, $(3)$ holds.
Example:
$\left.\binom{10000}{8687}\middle/\frac{1000^{8687}}{2^{8687}8687!}\right.=3.095040785$, but $\left.\binom{10000}{8688}\middle/\frac{1000^{8688}}{2^{8688}8688!}\right.=0.8127577101$