Prove that $s(z)=\frac{1}{1+e^{-z}}$ is always increasing I want to prove that $s(z)=\frac{1}{1+e^{-z}}$ is always increasing.
I know from previous work that taking the derivative and proving that it is always greater than $0$ is one way to prove $s(z)$ is always increasing.
How do I go about proving that the derivative $s'(z)=\frac{e^z}{(e^z+1)^2}$ is positive?
I know from plotting it graphically that I can see it is always positive, and from observing its limits I know the derivative approaches $0$ as $z\rightarrow -\infty,$ and $\infty$ as $z\rightarrow \infty$.
Is there a more rigorous proof I can use to show that $s'(z) > 0$ for all values of $z?$

$e^z$ is increasing, $e^{-z}$ is decreasing, $e^{-z}+1$ is decreasing and positive, $\dfrac1{e^{-z}+1}$ is increasing.

$$z_0<z_1\implies-z_0>-z_1\implies e^{-z_0}>e^{-z_1}\implies e^{-z_0}+1>e^{-z_1}+1>0 
\\\implies\frac1{e^{-z_0}+1}<\frac1{e^{-z_1}+1}.$$
