<s>[INST] Why is $(1+\frac{1}{n})^n < (1+\frac{1}{m})^{m+1}$? Why is it that
$$ 
\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{m}\right)^{m+1},
$$
for any natural numbers $m, n$? 
I have tried expanding using the binomial series and splitting into cases. I understand why it is trivially true when $m=n$ but I am not sure if there is a rigorous proof for other cases?
 [/INST] Both sequences
$$
a_n=\left(1+\frac{1}{n}\right)^n, \quad
b_n=\left(1+\frac{1}{n}\right)^{n+1}
$$
are monotonic. In particular, $a_n$ is increasing and $b_n$ is decreasing. 
Hence, if $m,n\in \mathbb N$ and $k=$max$\{m,n\}$, then
$$
a_n\le a_k\le b_k\le b_m.
$$
Why are these sequences monotonic?
We have
$$
\frac{a_{n+1}}{a_n}=\frac{(n+2)^{n+1}n^n}{(n+1)^{2n+1}}=\frac{n+2}{n+1}\cdot\left(\frac{n^2+2n}{(n+1)^2}\right)^n \\ =\frac{n+2}{n+1}\cdot\left(1-\frac{1}{(n+1)^2}\right)^n \ge \frac{n+2}{n+1}\cdot\left(1-\frac{n}{(n+1)^2}\right)=\frac{(n+2)(n^2+n+1)}{(n+1)^3}=\frac{n^3+3n^2+3n+2}{n^3+3n^2+3n+1}>1.
$$
In a similar fashion we get that $b_n$ is decreasing:
$$
\frac{b_n}{b_{n+1}}=\frac{\left(1+\frac{1}{n}\right)^{n+1}}{\left(1+\frac{1}{n+1}\right)^{n+2}}=\frac{(n+1)^{2n+3}}{n^{n+1}(n+2)^{n+2}}=\frac{n+1}{n+2}\cdot \left(\frac{n^2+2n+1}{n^2+2n}\right)^{n+1} \\ = \frac{n+1}{n+2}\cdot \left(1+\frac{1}{n^2+2n}\right)^{n+1}\ge \frac{n+1}{n+2}\cdot 
\left(1+\frac{n+1}{n^2+2n}\right) =\frac{(n+1)(n^2+3n+1)}{(n+2)(n^2+2n)}=\frac{n^3+4n^2+4n+1}{n^3+4n^2+4n}>1.
$$
</s>