How does $\sigma’(x)$ simplify to $\sigma(x)(1-\sigma(x))$?

Question

\begin{aligned} \sigma^{\prime}(x) &=\frac{\partial}{\partial x} \frac{1}{1+e^{-x}} \\ &=\frac{e^{-x}}{\left(1+e^{-x}\right)^{2}} \\ &=\frac{1}{1+e^{-x}} \cdot \frac{e^{-x}}{1+e^{-x}} \\ &=\sigma(x)(1-\sigma(x)) \end{aligned}

I do not understand what is the last step in this derivative. How can you factor out the last term?

\begin{aligned} \sigma(x)(1-\sigma(x)) \end{aligned}

Answer 1

$$\frac{e^{-x}}{1+e^{-x}} = \frac{(1+e^{-x})-1}{1+e^{-x}} = 1 - \frac{1}{1+e^{-x}} = 1- \sigma(x)$$