How to prove $(1+1/x)^x$ is increasing when $x>0$?

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.

It's enough to show $\ln(1+1/x)^x = x\ln(1+1/x)$ is increasing. Letting $h=1/x$, we can write the last expression as

$$\tag 1 \frac{\ln(1+h)-\ln 1}{h}.$$

We want to show $(1)$ increases as $h$ decreases. Now $(1)$ is just the slope of the line through $(1,\ln 1)$ and $(1+h,\ln(1+h)).$ And any concave function has the property that such slopes increase as $h$ decreases. Since $\ln x$ is concave, we're done.

$(1+1/x)^x$ is increasing equivalent to $x\cdot\ln(1+1/x)$ is increasing by taking natural logarithm. Which is equivalent to derivative being positive, which is showing that $h(x) = \ln(1+1/x) - 1/(1+x) > 0$.

And now notice $h'(x)=-1/[x\cdot(x+1)^2]$ is negative, so $h(x)$ is decreasing and minimal value is at $+\infty$, but $h(+\infty)=0$. So, $h(x) \ge 0$.