What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists?
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.
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...
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 $b_n >b_{n+1}$.