Inequality $\binom{2n}{n}\leq 4^n$ I would like to prove the following inequality, for $n=0,1,2,...$,   $$ \binom{2n}{n}\leq 4^n.$$ I already proved it by induction, and I'm looking for another proof. 
 A: Let $n \in \mathbb{N}.$ You may write
$$
4^n=2^{2n} =(1+1)^{2n}= \sum_{k=0}^{2n}\binom{2n}{k}\geq\binom{2n}{n}.
$$
A: $\displaystyle 4^n=(1+1)^{2n} =\sum_{k=0}^{2n}\binom{2n}{k}\geq\binom{2n}{n}$
A: A bit more can be proven with a bit more work. For $k\ge0$, we have the inequality
$$
\begin{align}
\left(\frac{k+\frac12}{k+1}\right)^2
&=\frac{k^2+k+\frac14}{k^2+2k+1}\\
&\le\frac{k+1}{k+2}\tag{1}
\end{align}
$$
because cross-multiplication gives $k^3+3k^2+\frac94k+\frac12\le k^3+3k^2+3k+1$. 
Using $(1)$ yields
$$
\begin{align}
\frac{\binom{2k+2}{k+1}}{\binom{2k}{k}}
&=4\frac{k+\frac12}{k+1}\\
&\le4\sqrt{\frac{k+1}{k+2}}\tag{2}
\end{align}
$$
Multiplying $(2)$ for $k=0$ to $k=n-1$, we get
$$
\boxed{\bbox[5px]{\displaystyle\binom{2n}{n}\le\frac{4^n}{\sqrt{n+1}}}}\tag{3}
$$

As Olivier Oloa comments, Stirling's Formula tells us that
$$
\lim_{n\to\infty}\binom{2n}{n}\frac{\sqrt{\pi n}}{4^n}=1\tag{4}
$$
In fact, using inequalities similar to $(2)$, in this answer, it is shown that
$$
\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi(n+\frac13)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}}}\tag{5}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{{2n \choose n}\leq 4^{n}:\ {\large ?}}$

\begin{align}
\binom{2n}{n}&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}
\end{align}

\begin{align}
\color{#66f}{\large\verts{\binom{2n}{n}}}&=
\verts{\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n} \over z^{n + 1}}
\,{\dd z \over 2\pi\ic}}
\\[5mm] \leq &
\oint_{\verts{z}\ =\ 1}\verts{\pars{1 + z}^{2n}}
\,{\verts{\dd z} \over 2\pi}<\pars{1 + 1}^{2n}=\color{#66f}{\LARGE 4^{n}}
\end{align}
