Lesser known derivations of well-known formulas and theorems What are some lesser known derivations of well-known formulas and theorems?
I ask because I recently found a new way to derive the quadratic formula which didn't involve completing the square as is commonly taught. Doing so I was wondering what other proofs and derivations for other formulas that have remained unknown to most people?
Whether it be because the proof is too complex or less pretty, I still find it insightful to see different ways to solve a problem. To me it makes me understand  proofs better, and thus also giving a better comprehension of it.

$$
\begin{align}
&\text{Given a quadratic function } f:\\[0.1em]
f &=ax^2+bx+c = a(x-r_1)(x-r_2) = ax^2-a(r_1+r_2)+ar_1r_2\\[0.1em]
a &= a,\enspace \frac{b}{a} = -(r_1+r_2),\enspace \frac{c}{a} = r_1r_2\\[1em]
f' &= 2ax+b, \enspace f'(x) = 0 \Rightarrow x = -\frac{b}{2a}\\[0.2em]
\text{This is an}& \text{ extremum of } f \text{, and is equidistant from each root } r_1, \enspace r_2 \text{ as shown:}\\[0.4em]
\frac{b}{a} &= -(r_1+r_2) \iff -\frac{b}{2a} = \frac{r_1+r_2}{2} \\[1em]
\Rightarrow \enspace &\text{The roots are of the form } r= -\frac{b}{2a}\pm d\\[1em]
\frac{c}{a} = r_1r_2 &= (-\frac{b}{2a}+d)(-\frac{b}{2a}-d) = \frac{b^2}{4a^2}-d^2\\[0.2em]
\Rightarrow\enspace& d^2 = \frac{b^2}{4a^2}-\frac{c}{a} = \frac{b^2}{4a^2}-\frac{4ac}{4a^2} = \frac{b^2-4ac}{4a^2}\\[0.2em]
\Rightarrow\enspace&d = \pm\frac{\sqrt{b^2-4ac}}{2a}\\[1em]
\text{Which yields }& r = -\frac{b}{2a}\pm d = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\enspace\square
\end{align}
$$


The proof in
N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58
(1951), 563
that
$(1+1/n)^n
\to e$
which uses the arithmetic-geometric mean
inequality (AGMI),
 which we will use in the
form 
$((v_1+v_2+...v_n)/n)^n > v_1v_2...v_n$
 (all $v_i$ positive) 
with equality if and only if all the $v_i$ are
equal. 
Consider $n$ values of $1+1/n$
and $1$ value of $1$. 
By the AGMI, 
$((n+2)/(n+1))^{n+1} > (1+1/n)^n
$, or 
$(1+1/(n+1))^{n+1} > (1+1/n)^n$.
Consider $n$ values of $1-1/n$
 and $1$ value of $1$. 
By the AGMI,
$(n/(n+1))^{n+1} > (1-1/n)^n$
 or
 $(1+1/n)^{n+1} < (1+1/(n-1))^n
$.
Therefore
$(1+1/n)^n$
is increasing
and
$(1+1/n)^{n+1}$
is decreasing.
Since
$0
< (1+1/n)^{n+1}-(1+1/n)^n
= (1+1/n)^{n}/n
<4/n
$,
the two sequences converges to
a common limit.
To show that
$(1+1/n)^{n}
< 4
$,
note that
$(1+1/n)^{n}
<(1+1/n)^{n+1}
\le 2^2
$
 (with 
$n=1)
=4
$,
