Proving $\left(1+\frac{1}{m}\right)^m < \left(1+\frac{1}{n}\right)^n$ Let $m,n\in \mathbb{N}$. If $m > n$ show that
$$\left(1+\frac{1}{m}\right)^m > \left(1+\frac{1}{n}\right)^n$$
My works: I tried to show if $g(x)=\left(1+\frac{1}{x}\right)^x$ then $g'(x) > 0$.
\begin{align}
g'(x) &= \frac{d e^ { x \ln(1+\frac{1}{x}) }} {dx} \\
&= e^{x\ln(1+1/x)} \left(\ln(1+\frac{1}{x}) - \frac{1}{1+x} \right) >0
\end{align}

Normally the proof goes something like, let $a_n=(1+\frac{1}{n})^n$ for all $n\in\mathbb Z^+$. Then, $$\frac{a_{n+1}}{a_n}=\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^n}=\frac{(\frac{n+2}{n+1})^{n+1}}{(\frac{n+1}{n})^n}=\frac{(n+2)^{n+1}n^n}{(n+1)^n(n+1)^{n+1}}=\frac{(n^2+2n)^n}{(n+1)^{2n}}\frac{n+2}{n+1}=\frac{((n+1)^2-1)^n}{((n+1)^2)^n}\frac{n+2}{n+1}=\bigg(1-\frac{1}{(n+1)^2}\bigg)^n\frac{n+2}{n+1}\geq\bigg(1-\frac{n}{(n+1)^2}\bigg)\frac{n+2}{n+1}=\frac{n^2+n+1}{(n+1)^2}\frac{n+2}{n+1}=\frac{(n^2+n+1)(n+2)}{(n+1)^3}=\frac{n^3+3n^2+3n+2}{n^3+3n^2+3n+1}> 1$$ 
(the first inequality that appears above comes from Bernoulli's inequality which can be proven by simple induction as seen in the link)  
So, $\frac{a_{n+1}}{a_n}>1\implies a_{n+1}>a_n$. This means if $m$ is any integer greater than $n$, $$a_m>a_{m-1}>a_{m-2}>...>a_{n+1}>a_n$$
giving us $a_m>a_n$, i.e., $(1+\frac{1}{m})^m>(1+\frac{1}{n})^n$
