I was playing around, and decided to see if I could come up with a proof that $$\lim_{n \to \infty} (1+1/n)^n$$ exists that was as elementary as possible in that it uses only Bernoulli's inequality.
This is probably not new, but I haven't seen it before. If it is already here, so be it.
Here it is.
Let $p_n =(1+1/n)^n $.
\begin{align} \frac{p_{n+1}}{p_n} &=\frac{(1+1/(n+1))^{n+1}}{(1+1/n)^n}\\ &=\frac{((n+2)/(n+1))^{n+1}}{((n+1)/n)^n}\\ &=\frac{(n+2)^{n+1}n^n}{(n+1)^{2n+1}}\\ &=\frac{(n(n+2))^{n+1}}{n(n+1)^{2n+1}}\\ &=\frac{(n+1)(n(n+2))^{n+1}}{n(n+1)^{2n+2}}\\ &=\frac{n+1}{n}\left(\frac{n(n+2)}{(n+1)^{2}}\right)^{n+1}\\ &=\frac{n+1}{n}\left(\frac{n^2+2n}{n^2+2n+1}\right)^{n+1}\\ \end{align}
By the version of Bernoulli's inequality which states $(1-\frac1{m})^n > 1-\frac{n}{m}$ if $m > n$,
\begin{align} \left(\frac{n^2+2n}{n^2+2n+1}\right)^{n+1} &=\left(1-\frac{1}{n^2+2n+1}\right)^{n+1}\\ &>\left(1-\frac{n+1}{n^2+2n+1}\right)\\ &=\left(1-\frac{1}{n+1}\right)\\ &=\frac{n}{n+1} \end{align}
so $$\frac{p_{n+1}}{p_n} \ >\frac{n+1}{n}\frac{n}{n+1} = 1. $$
Therefore, $p_n$ is an increasing sequence.
Going the other way, by Bernoulli's inequality,
\begin{align} \left(\frac{n^2+2n+1}{n^2+2n}\right)^{n+1} &=\left(1+\frac{1}{n^2+2n}\right)^{n+1}\\ & > 1+\frac{n+1}{n^2+2n}\\ &=\frac{n^2+3n+1}{n^2+2n} \end{align}
so
\begin{align} \frac{p_{n+1}}{p_n} &<\frac{n+1}{n}\frac{n^2+2n}{n^2+3n+1}\\ &=\frac{(n+1)(n+2)}{n^2+3n+1}\\ &=\frac{n^2+3n+2}{n^2+3n+1}\\ &<\frac{n^2+3n+2}{n^2+3n}\\ &=\frac{(n+1)(n+2)}{n(n+3)} \end{align}
and
\begin{align} \frac{p_{n+m}}{p_n} &=\prod_{k=0}^{m-1} \frac{p_{n+k+1}}{p_{n+k}}\\ &=\prod_{k=0}^{m-1}\frac{(n+1)(n+2)}{n(n+3)}\\ &=\prod_{k=0}^{m-1}\frac{n+1}{n}\prod_{k=0}^{m-1}\frac{n+2}{n+3}\\ &=\frac{n+m}{n}\frac{n+2}{n+m+2}\\ &=\frac{n^2+n(m+2)+2m}{n^2+n(m+2)}\\ &=1+\frac{2m}{n^2+n(m+2)}\\ &<1+\frac{2m}{nm}\\ &=1+\frac{2}{n} \end{align}
Therefore $p_n$ is bounded above.
Since it is an increasing sequence, it has a limit.
Note: Here is a proof that $(1-\frac1{m})^n >1-\frac{n}{m}$ if $2 \le n \le m$.
It is true for $n=2$, since \begin{align} (1-\frac1{m})^2 &=1-\frac{2}{m}+\frac1{m^2}\\ & > 1-\frac{2}{m} \end{align}
If it is true for $n$, and $n < m$, then
\begin{align} (1-\frac1{m})^{n+1} &=(1-\frac1{m})^n (1-\frac1{m}) \\ &\ge(1-\frac{n}{m}) (1-\frac1{m}) \\ &=1-\frac{n+1}{m}+\frac{n}{m^2} \\ &>1-\frac{n+1}{m} \\ \end{align}
I find this an interesting proof by induction since it holds for any $m$ for a range of $n$ values, rather than all $n$.
