Let $F(x)=(1+\frac{1}{x})^x$.
How do we prove $F(x)$ is increasing when $x>0$?
Let $F(x)=(1+\frac{1}{x})^x$.
How do we prove $F(x)$ is increasing when $x>0$?
$f(t):=\frac{1}{t}\log(1+t)$ is decreasing for $t>0$, because it is smooth and its derivative is negative.
Its derivative is f'(t)=-\frac{1}{t^2}g(t), where $g(t):=\log(1+t)-\frac{t}{1+t}.$
But, $g(t)>0$ for $t>0$, in fact, $\lim_{t\to 0}\ g(t)=0$ and $g$ is increasing.
$g$ is increasing because it is smooth and its derivative is positive: g'(t)=\frac{t}{(1+t)^2}.
There's an elementary approach for rational $x$. It suffices to prove that $\left( 1+\frac{m}{n} \right)^n < \left( 1+\frac{m}{n+1} \right)^{n+1}$ for $m,n$ positive integers. Whenever $0 \leq a < b$, we have $\frac{b^{n+1} - a^{n+1}}{b-a} = \sum_{k=0}^n a^{n-k}b^k < (n+1)b^n$ which rearranges to $[(n+1)a - nb] \cdot b^n < a^{n+1}.$
Substituting $a = 1+m/(n+1)$ and $b = 1 +m/n$ into the above, the term in square braces (miraculously) reduces to $1$ and we get the desired bound. This is adapted from Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger.
Let $h(x) = \mathrm e^{-1/(1+x)}(1+1/x)$ and note that for $x > 0$, $ h(x) = \mathrm e^{-1/(1+x)}\cdot\frac{x+1}{x} > \left(1-\frac{1}{1+x}\right)\frac{x+1}{x} = 1. $
Now, let $g(x) = \log F(x)$ and note that g'(x) = \log(1+1/x) - \frac{1}{1+x} = \log h(x) > \log 1 = 0\>, and so we are done.