# Proof of $e^{-x}/(1+e^{-x})^2$ is equal to $e^x/(1+e^x)^2$

I am stuck at proving that $e^x/(1+e^x)^2$ is equal to $e^{-x}/(1+e^{-x})^2$ as it is provided here

$$\frac{e^x}{(1+e^x)^2} = \frac{e^x}{(e^x(e^{-x}+1))^2} = \frac{e^x}{e^{2x}(e^{-x}+1)^2}= \frac{e^{-x}}{(1+e^{-x})^2}$$
Multiply numerator and denominator by $e^{-2x}$:
$$\frac{e^x}{(1+e^x)^2}=\frac{e^xe^{-2x}}{(1+e^x)^2e^{-2x}}=\frac{e^{-x}}{((1+e^x)e^{-x})^2}=\frac{e^{-x}}{(e^{-x}+1)^2}$$
It is enough to prove that $f(x)=\frac{e^x}{(1+e^x)^2}$ is an even function, and $$f(x) = \frac{e^{-x}}{(e^{-x/2})^2}\cdot\frac{e^x}{(1+e^x)^2}=\frac{1}{\left(e^{x/2}+e^{-x/2}\right)^2} = \frac{1}{4\cosh^2\frac{x}{2}}$$ is quite trivially an even function.
Consider $$1+\frac{1}{a}=\frac{1+a}{a}$$ and then $$\frac{\dfrac{1}{a}}{\left(1+\dfrac{1}{a}\right)^{\!2}}= \frac{\dfrac{1}{a}}{\dfrac{(1+a)^2}{a^2}}= \frac{1}{a}\frac{a^2}{(1+a)^2}=\frac{a}{(1+a)^2}$$ Now set $a=e^x$.