Exponential simplification help Why is it true that
$$
\frac{exp(x)}{1+exp(x)} = \frac{1}{exp(-x) + 1}
$$
Why did we negate $x$?
I tried to divide the left-hand side with $exp(x)$ and got
$$
\frac{exp(x)}{exp(x) + (exp(x))^2}
$$
but I cannot get forward.
$$\dfrac{\exp(x)}{1+\exp(x)} = \dfrac{\frac1{\exp(x)}\exp(x)}{\frac1{\exp(x)}(1+\exp(x))} = \dfrac{1}{\frac1{\exp(x)}+1} =  \dfrac{1}{{\exp(-x)}+1}$$
Perhaps you should tell us what $\exp(x)$ is defined as.
The simplest explanation is that you know there is a number $e$ and you know what raising a base to a power is and $\exp(x)$ is simply defined as $\exp(x)= e^x$.  And therefore $\exp(-x) = e^{-x}= \frac 1{e^x} = \frac 1{exp(x)}$ (and we know that $e^{-x}= \frac 1{e^x}$ because $e^{-x}e^{x} = e^{-x+x} = e^0 = 1$).
But that assume you know what raising numbers to a power means.  We usually use the concept of $\exp(x)$ to define what raising a number to a power is by $b^x = \exp(x\cdot \ln b)$ where $\ln b$ is defined as $\int_1^b \frac 1t dt$ and $\exp(x) = \ln^{inverse}(x)$ (i.e. $\exp(x)$ is the value $k$ so that $\ln (k) = x$.
Then its a matter of proving that $\exp(x+y) = \exp x\cdot \exp y$.  If we know that then we have to know that $\exp(x)\exp(-x) = \exp(x-x)=\exp(0) = 1$ and therefore $\exp(-x) =\frac 1{\exp(x)}$.
I may provide a proof of that in a few hours.