TITLE: Prove that $e>2$ geometrically.
QUESTION [18 upvotes]: Q: Prove that $e>2$ geometrically.
Attempt: I only know one formal definition of $e$ that is $\lim_\limits{n\to\infty} (1+\frac{1}{n})^n=e$. I could somehow understand that this is somehow related to rotation in the complex plane.
$$e^{i\theta}=\cos \theta + i \sin \theta$$
Hence we have $$e^{i\pi}=-1$$
But how can I bring out the value of $e$ when I am showing this rotation in a geometrical figure?
Any hints are appreciated.
EDIT: As per the comments, I am making a small addition to the question which will not affect the existing answers. It is that, as a definition of $e$, one can use any definition which does not use the fact $2<e<3$.

REPLY [24 votes]: In this image

we see that
$$
\color{#00A000}{1}+\color{#C000C0}{x}\le\left(1+\frac x2\right)^2
$$
Therefore,
$$
\begin{align}
1+1
&\le\left(1+\frac12\right)^2\\
&\le\left(1+\frac14\right)^4\\
&\le\left(1+\frac18\right)^8\\
&\dots\\
&\le\lim_{n\to\infty}\left(1+\frac1{2^n}\right)^{\large2^n}\\[9pt]
&=e
\end{align}
$$