${\displaystyle {\begin{aligned}\left(1+{\frac {r}{n}}\right)^{n}&=\sum _{k=0}^{n}{n \choose k}{\frac {r^{k}}{n^{k}}}\\&=\sum _{k=0}^{n}{\frac {n!r^{k}}{n^{k}k!\left(n-k\right)!}}\\&=1+\sum _{k=1}^{n}{\frac {n(n-1)\cdots (n-k+1)(n-k)!r^{k}}{n^{k}k!\left(n-k\right)!}}\\&=1+\sum _{k=1}^{n}{\frac {n(n-1)\cdots (n-k+1)r^{k}}{n^{k}k!}}\\&\leq 1+\sum _{k=1}^{n}{\frac {r^{k}}{k!}}\\\end{aligned}}}$

At this stage ${\displaystyle 2^{k-1}}$ was shown to be less than or equal to ${\displaystyle k!}$ which is correct but if this is done the conclusion of the proof is:

${\displaystyle \left(1+{\frac {r}{n}}\right)^{n}\leq 1+r{\frac {1-\left({\frac {r}{2}}\right)^{n}}{1-{\frac {r}{2}}}}}$

${\displaystyle \left({\frac {r}{2}}\right)^{n}}$ will increase without bound if ${\displaystyle r>2}$, so this method does not prove the general case that ${\displaystyle \left(1+{\frac {r}{n}}\right)^{n}}$ is bounded. (When ${\displaystyle r=2}$ there is a divide-by-zero problem as well.)

A possible although unwieldy solution is to replace k! with something other than ${\displaystyle 2^{k-1}}$. I propose a complete solution as follows:

(using ${\displaystyle \left\lceil n\right\rceil }$ as the ceiling function):

${\displaystyle k!\geq \left\lceil r+1\right\rceil ^{k-\left\lceil r\right\rceil }}$

therefore:

${\displaystyle {\begin{aligned}\left(1+{\frac {r}{n}}\right)^{n}&\leq 1+\sum _{k=1}^{n}r^{\left\lceil r\right\rceil }\left({\frac {r}{\left\lceil r+1\right\rceil }}\right)^{k-\left\lceil r\right\rceil }\\&=1+r^{\left\lceil r\right\rceil }\sum _{k=1}^{n}\left({\frac {r}{\left\lceil r+1\right\rceil }}\right)^{k-\left\lceil r\right\rceil }\end{aligned}}}$

${\displaystyle \sum _{k=1}^{n}\left({\frac {r}{\left\lceil r+1\right\rceil }}\right)^{k-\left\lceil r\right\rceil }}$ is a GP, so using ${\displaystyle S_{n}={\frac {T_{0}(1-r^{n})}{1-r}}}$ with ${\displaystyle n}$ as the number of terms and ${\displaystyle r}$ as the growth per term:

${\displaystyle \sum _{k=1}^{n}\left({\frac {r}{\left\lceil r+1\right\rceil }}\right)^{k-\left\lceil r\right\rceil }={\frac {\left({\frac {r}{\left\lceil r+1\right\rceil }}\right)^{1-\left\lceil r\right\rceil }\left(1-\left({\frac {r}{\left\lceil r+1\right\rceil }}\right)^{n}\right)}{1-{\frac {r}{\left\lceil r+1\right\rceil }}}}}$

But for all ${\displaystyle r>0}$:

${\displaystyle 0<{\frac {r}{\left\lceil r+1\right\rceil }}<1}$

So ${\displaystyle 0<1-\left({\frac {r}{\left\lceil r+1\right\rceil }}\right)^{n}<1}$

therefore:

${\displaystyle \left(1+{\frac {r}{n}}\right)^{n}\leq 1+{\frac {r^{\left\lceil r\right\rceil }\left({\frac {r}{\left\lceil r+1\right\rceil }}\right)^{1-\left\lceil r\right\rceil }}{1-{\frac {r}{\left\lceil r+1\right\rceil }}}}}$

which is a bound independent of ${\displaystyle n}$