Proof that $(1+1/x)^x$ is monotonic increasing [duplicate]

How does one prove that $(1+\frac{1}{x})^x$ is monotonic increasing for any $x \in [1,\infty)$?

Thanks a million!

For $x\in \mathbb{N}$ this result is very obvious. $$(\frac{n.\frac{n+1}{n}+1}{n+1})\geq((\frac{n+1}{n})^n.(1)^{1})^{(\frac{1}{n+1})}$$ Using A.M. G.M. inequality(i.e A.M$\geq$G.M.) $$\Rightarrow\frac{(n+1)+1}{n+1}\geq(\frac{n+1}{n})^{\frac{n}{n+1}}$$ $$\Rightarrow(1+\frac{1}{n+1})^{n+1}\geq(1+\frac{1}{n})^{n}$$

$$f(x) = \left ( 1+ \frac{1}{x} \right )^x$$

$$\log{f(x)} = x \log{\left( 1+ \frac{1}{x} \right )}$$

$$\frac{d}{dx}\log{f(x)} = \log{\left( 1+ \frac{1}{x} \right )} + \frac{x}{1+\frac{1}{x}}\left (\frac{-1}{x^2} \right ) = \log{\left( 1+ \frac{1}{x} \right )} - \frac{1/x}{1+ \frac{1}{x}}$$

or

$$\frac{d}{dx}\log{f(x)} = \log{\left( 1+ \frac{1}{x} \right )} + \frac{1}{1+ \frac{1}{x}} - 1 $$

Because $y \log{y} \ge y-1 \; \forall y \ge 1$, we may say that, $\forall x \ge 1$:

$$\frac{d}{dx}\log{f(x)} =\log{\left( 1+ \frac{1}{x} \right )} + \frac{1}{1+ \frac{1}{x}} - 1 \ge 0$$

and $f(x)$ is monotone increasing for all $x \ge 1$, as $\log$ is monotone increasing.

Hint: Consider the sign of the derivative!