What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists? Here is my candidate for 
the most elementary proof that $\lim_{n \to \infty}(1+1/n)^n $ exists.
I would be interested in seeing others.
$***$ Added after some comments:
I prove here by very elementary means that the limit exists.
Calling the limit "$e$" names it.
$***$
It only needs Bernoulli's inequality (BI)
in the form
$(1+x)^n \ge 1+nx$
for $x > -1$ and
$n$ a positive integer,
with equality only if
$x = 0$ or $n = 1$.
This is easily proved by induction:
It is true for $n=1$,
and $(1+x)^{n+1}
= (1+x)(1+x)^n
\ge (1+x)(1+nx)
= 1+(n+1)x+nx^2
\ge 1+(n+1)x
$.
(If $-1 < x < 0$,
if $1+nx \ge 0$,
the above proof goes through, and if
$1+nx < 0$,
$1+mx < 0$ for all $m \ge n$
so certainly $(1+x)^m > 1+mx$.)
This proof originally appeared in
N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563
and uses the
arithmetic-geometric mean inequality (AGMI)
in the form
$\big(\sum_{i=1}^n v_i/n\big)^n \ge \prod_{i=1}^n v_i$
 (all $v_i$ positive) with equality if and only if all the $v_i$ are equal.
Let $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$. 
We will prove that $a_n$ is an increasing sequence
 and $b_n$ is an decreasing sequence. 
Since $a_n < b_n$, this implies, 
for any positive integers $n$ and $m$ with $m < n$
 that $a_m < a_n < b_n < b_m$.
For $a_n$, consider n values of $1+1/n$ and $1$ value of $1$. 
Since their sum is $n+2$
and their product is $(1+1/n)^n$,
by the AGMI, 
$((n+2)/(n+1))^{n+1} > (1+1/n)^n$,
 or $(1+1/(n+1))^{n+1} > (1+1/n)^n$, 
or $a_{n+1} > a_n$. 
For $b_n$, consider $n$ values of $1-1/n$ and $1$ value of $1$. 
Since their sum is $n$
and their product is $(1-1/n)^n$,
by the AGMI, 
$(n/(n+1))^{n+1} > (1-1/n)^n$
 or $(1+1/n)^{n+1} < (1+1/(n-1))^n$,
 or $b_n > b_{n+1}$.
Since $b_n-a_n 
= (1+1/n)^{n+1} - (1+1/n)^n
= (1+1/n)^n(1/n)
=a_n/n
$
and every $a_n$ is less than any $b_n$
and $b_3 = (1+1/3)^4 = 256/81 < 4$,
$b_n-a_n < 4/n$,
so $b_n$ and $a_n$ converge to a common limit.
These proofs do not seem to be really elementary,
 since they use the AGMI. 
However, they use a special form of the AGMI, 
where all but one of the values are the same, 
and this will now be shown to be implied by BI,
 and thus be truly elementary. 
Suppose we have $n-1$ values of $u$ and $1$ value of $v$ with $u$ and $v$ positive. 
The AGMI for these values is 
$(((n-1)u+v)/n)^n \ge u^{n-1}v$ 
with equality if and only if $u = v$. 
We will now show that this is implied by BI:
$(((n-1)u+v)/n)^n \ge u^{n-1}v$
 is the same as 
$(u+(v-u)/n)^n \ge u^n(v/u)$. 
Dividing by $u^n$, 
this is equivalent to 
$(1+(v/u-1)/n)^n \ge v/u$. 
By BI, 
since $(v/u-1)/n > -1/n > -1$,
$(1+(v/u-1)/n)^n \ge 1+n((v/u-1)/n) = v/u$ 
with equality only if $n=1$ or $v/u-1 = 0$.
Thus BI implies this version of the AGMI.
 A: Could this be ?
$$\begin{align} \frac{d}{dx} \ln x &= \lim_{h \to 0}\; \frac{\ln(x + h) − \ln x}{h} \\
                   &= \lim_{h \to 0} \;\frac{\ln(1 + \frac{h}{x})}{h} \\
                   &= \lim_{h \to 0} \;\ln\left((1 + \frac{h}{x})^{\frac{1}{h}}\right) \\
\end{align}$$
let $h=\frac xc$
$$\begin{align} \frac{d}{dx} \ln x &= \lim_{c \to \infty} \;\ln\;\left((1 + \frac{1}{c})^{\frac{c}{x}}\right) \\
                                   &= \frac{1}{x} \;\lim_{c \to \infty}\; \ln\;\left((1 + \frac{1}{c})^{c}\right) \\
                                   &= \frac{1}{x} \;\ln\;\left(\lim_{c \to \infty}\;(1 + \frac{1}{c})^{c}\right)
\end{align}$$
Since $\frac{d}{dx} \ln x = 1/x$
$$\begin{align} \frac{1}{x} &= \frac{1}{x} \;\ln\;\left(\lim_{c \to \infty}\;(1 + \frac{1}{c})^{c}\right) \\
                          1 &= \ln\;\left(\lim_{c \to \infty}\;(1 + \frac{1}{c})^{c}\right) \\
                          e &= \lim_{c \to \infty}\;\left(1 + \frac{1}{c}\right)^{c}
\end{align}$$
That's not very rigorous but well...
A: We can prove directly from BI at the expense of some algebra.
$(1+(v/u-1)/(n+1))^{n+1} > v/u$ with $u=1+1/n$ and $v=1$ is
$$\frac{1}{1+\frac{1}{n}} < \left(1+\frac{\frac{1}{1+\frac{1}{n}}-1}{n+1}\right)^{n+1} = \left( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right)^{n+1}$$
which gives $a_{n+1} >a_n$ in Marty's notation. With $u=1-1/n$ and $v=1$
$$\frac{1}{1-\frac{1}{n}} < \left(1+\frac{\frac{1}{1-\frac{1}{n}}-1}{n+1}\right)^{n+1} = \left( \frac{1+\frac{1}{n-1}}{1+\frac{1}{n}} \right)^{n+1}$$
Since $(1-\frac{1}{n})(1+\frac{1}{n-1}) = 1$
this gives $b_n >b_{n+1}$.
A: If you just need to show that the $\lim\limits_{n\to \infty}\left(1 + \frac{1}{n}\right)^{n}$ exists, you can just show that it is an increasing sequence which is bounded above. To show that it is increasing you just need binomial theorem (an elementary result) and to prove that it is bounded, you can expand by binomial and then compare with the geometric series with ratio $\frac{1}{2}$.
Then you can define $e$ to be the limit of this sequence. And without too much effort, one may also show that
$$ e = \lim\limits_{n\to \infty}\sum\limits_{k=0}^{n}\frac{1}{k!}$$.
You will find all the details in the Principles of Mathematical Analysis by Walter Rudin.
