Ask about beautiful properties of $e$ One of students asked me about "some beautiful properties (or relation) of $e$". Then I list like below 
\begin{align}
& e \equiv \lim_{x \to \infty} \left(1+\frac{1}{x} \right)^x\\[10pt]
& e = \sum_{k=0}^\infty \frac{1}{k!}\\[10pt]
& \frac{d}{dx} (e^x) = e^x\\[10pt]
& e^{ix} = \cos x  + i \sin x  \quad \text{(Euler)}\\[10pt]
& e^{i \pi} + 1 = 0
\end{align} After this, he asked me for more relation or properties. I said 
I'll think and answer ...
Now I want help to add some relation, properties, or visual things (like proof without words)
Please help me to add something more. Thanks in advance.
***The class was math. 1. engineering
 A: My favourite is about derangements.
$n$ people write their name on different envelopes and put the envelopes in a box.
If they pick a random envelope back from the box, the probability that no one picks an envelope with its name is $\approx\frac{1}{e}$.

A number-theoretic one: if $\{p_k\}_{k\geq 1}$ is the sequence of primes,
$$ \prod_{k=1}^{n} p_k = e^{n+o(n)}.$$
A: Here  are some more   relations which might be pleasing.

From section 1.3 of Mathematical Constants by S.R. Finch:
  
  
*
  
*A    Wallis-like  infinite   product is
  \begin{align*}
e=\frac{2}{1}\cdot\left(\frac{4}{3}\right)^{\frac{1}{2}}
\cdot\left(\frac{6\cdot 8}{5\cdot 7}\right)^{\frac{1}{4}}
\cdot\left(\frac{10\cdot 12\cdot 14\cdot 16}{9\cdot 11\cdot 13\cdot 15}\right)^{\frac{1}{8}}\cdots
\end{align*}
  
*From Stirling's formula we derive
  \begin{align*}
e=\lim_{n\rightarrow  \infty}\frac{n}{(n!)^{\frac{1}{n}}}
\end{align*}
  
*Another continued fraction representation is
  \begin{align*}
e&=2+\frac{\left.2\right|}{\left|2\right.}
+\frac{\left.3\right|}{\left|3\right.}
+\frac{\left.4\right|}{\left|4\right.}
+\frac{\left.5\right|}{\left|5\right.}+\cdots\\
&=2+\frac{2}{2+\frac{3}{3+\frac{4}{4+\frac{5}{5+\cdots}}}}
\end{align*}

$$ $$

In section Intriguing Results in Real Infinite Series by D.D. Bonar and M.J. Khoury we  find
  
  
*
  
*Gem 89 (American Math Monthly 42:2 pp. 111-112)
  
  
  \begin{align*}
e=\frac{1}{5}\left(\frac{1^2}{0!}+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\cdots\right)
\end{align*}

A: Here is the (simple) continued fraction for $e-1.$ The pattern continues forever, 1,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,...
To get $e$ itself add one, this is the same as changing the first $1$ to a $2$
$$  
\begin{array}{ccccccccccccccccccccccc}
 & & 1 & & 1 & & 2 & & 1 & & 1 & & 4 & & 1 & & 1 & & 6 &   \\
\frac{0}{1} & \frac{1}{0} & & \frac{1}{1} &  &  \frac{2}{1} & &  \frac{5}{3}  & & \frac{7}{4} & &  \frac{12}{7} & &   \frac{55}{32}  & &   \frac{67}{39}  & &   \frac{122}{71}   & &   \frac{799}{465}  \\
\end{array}
$$
A: Let's say that a number is cut into equal parts and then multiply those parts together, for example take number $20$ and divide it by $4$ ($20/4=5$), as result we have:
$5\times5\times5\times5=625$
The question here is to find the number $n\in \mathrm{N}$ such that is maximizes 
$r = \arg \max_{n,m} \prod_{i=1}^{n}m$
where
$m=(\mathrm{your\:choice\:of\:number})/n$
We can show that the number $r$ is maximized when $m\approx e$. In the case of choosing $20$ we have
$20/7\approx e$
so 
$r=(2.8)^7$
A: One of my favourites is the following property of $e$.
Write down a random number between $0$ and $1$. Write down another one and add it to the previous one. If the total exceeds $1$, stop. Otherwise keep adding such random numbers until the total exceeds $1$ and stop.
The expected number of such random numbers is $e$.
