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?
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: $$ \frac{b_n}{b_{n+1}}=\frac{\left(1+\frac{1}{n}\right)^{n+1}}{\left(1+\frac{1}{n+1}\right)^{n+2}}=\frac{(n+1)^{2n+3}}{n^{n+1}(n+2)^{n+2}}=\frac{n+1}{n+2}\cdot \left(\frac{n^2+2n+1}{n^2+2n}\right)^{n+1} \\ = \frac{n+1}{n+2}\cdot \left(1+\frac{1}{n^2+2n}\right)^{n+1}\ge \frac{n+1}{n+2}\cdot \left(1+\frac{n+1}{n^2+2n}\right) =\frac{(n+1)(n^2+3n+1)}{(n+2)(n^2+2n)}=\frac{n^3+4n^2+4n+1}{n^3+4n^2+4n}>1. $$
The inequality holding for all naturals implies that
$$\left(1+\frac{1}{n}\right)^n < L\le U < \left(1+\frac{1}{m}\right)^{m+1},$$
where $U=\inf_m\left(1+\dfrac{1}{m}\right)^{m+1}, L=\sup_n\left(1+\dfrac{1}{n}\right)^n $. There may not be any overlap between the values in the LHS and RHS.
Indeed for all real $x>0$, $$\left(1+\frac1x\right)^x<e$$ and $$\left(1+\frac1x\right)^{x+1}>e.$$
Both functions are monotonic and their limit $x\to\infty$ is $e$.
Monotonicity follows from the inequalities
$$\frac1{x+1}<\log\left(1+\frac1x\right)<\frac1x$$ which imply positive and negative derivatives respectively. These inequalities are supported by inequalities between the derivatives
$$-\frac1{(x+1)^2}>-\frac1{x(x+1)}>-\frac1{x^2}$$ and the common value $0$ at infinity.
For a simple answer to why is $\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{m}\right)^{m+1}$? -- it is because of the inequality of geometric and arithmetic means. In particular
$a_n = \big(1 + \frac{1}{n}\big)^n \lt \big(1 + \frac{1}{n+1}\big)^{n+1} = a_{n+1}$
or equivalently, taking n+1 th roots
$a_{n}^\frac{1}{n+1}= \big(1 + \frac{1}{n}\big)^\frac{n}{n+1} = \big(1 + \frac{1}{n}\big)^\frac{n}{n+1}\cdot 1^\frac{1}{n+1} \lt \frac{n}{n+1}\big(1 + \frac{1}{n}\big) + \frac{1}{n+1}\big(1\big) = 1 + \frac{1}{n+1} = a_{n+1}^\frac{1}{n+1}$
by $\text{GM}\leq \text{AM}$ (which holds with equality iff all items in the geometric mean are identical)
and by the same manipulation we get $c_r = \big(1 - \frac{1}{r}\big)^r \lt \big(1 + \frac{1}{r+1}\big)^{r+1} = c_{r+1}$
so $a_n$ and $c_r$ are strictly monotone increasing
For the question posed: by monotone behavior of $a_n$, if $n \leq m$ then we automatically have
$(1+\frac{1}{n})^n \leq (1+\frac{1}{m})^{m}\lt (1+\frac{1}{m})^{m}\cdot (1+\frac{1}{m}) = (1+\frac{1}{m})^{m+1}$
it remains to consider the case of $m \lt n$. In this case we can divide out $(1+\frac{1}{m})^{m+1}$ and prove the equivalent
$\left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{m+1}\right)^{m+1} = \left(1+\frac{1}{n}\right)^n \left(\frac{m}{m+1}\right)^{m+1}=\left(1+\frac{1}{n}\right)^n \left(1+\frac{1}{m}\right)^{-(m+1)}\lt 1 $
but
$\left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{m+1}\right)^{m+1} \leq \left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{n}\right)^{n}\lt 1$
by the monotone behavior of $c_r$ and $\text{GM}\leq \text{AM}$ which proves the claim
note: if you don't like splitting this into 2 cases, then the claim is proven directly by considering monotone behavior in $a_n$ and $c_r$, and for any $m,n$ select natural number $k \gt \max(m+1,n)$ and the direct proof is
$\left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{m+1}\right)^{m+1} \lt \left(1+\frac{1}{k}\right)^k \left(1-\frac{1}{k}\right)^{k}\lt 1$
and of course
$\left(1+\frac{1}{k}\right)^k \left(1-\frac{1}{k}\right)^{k}\lt 1$ is equivalent to proving the claim with 2kth roots of each side but
$\left(1+\frac{1}{k}\right)^\frac{1}{2} \left(1-\frac{1}{k}\right)^\frac{1}{2}\lt \frac{1}{2}\left(1+\frac{1}{k}\right)+ \frac{1}{2}\left(1-\frac{1}{k}\right) = 1$ by $\text{GM}\leq \text{AM}$
