Why is $(1+\frac{1}{n})^n < (1+\frac{1}{m})^{m+1}$?
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.