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?

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 $\text{GM}\leq \text{AM}$ 
