Prove $0.9999^{\!101}<0.99<0.9999^{\!100}$ Prove
$$0.9999^{\!101}<0.99<0.9999^{\!100}$$
I think its original idea is 
$$(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})} \tag{I can't prove!}$$
For $x=100^{\!-1}\,\therefore\,(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})}\,\therefore\,0.99^{\!-99}<1.01^{\!100}$.
Furthermore $0.99^{\!-99}\times0.99^{\!100}=0.99<1.01^{\!100}\times0.99^{\!100}=0.9999^{\!100}$.
For $x=-100^{\!-1}\,\therefore\,(1-x)^{(1-\frac{1}{x})}<(1+x)^{(\frac{1}{x})}\,\therefore\,1.01^{\!101}<0.99^{\!-100}$.
Furthermore $1.01^{\!101}\times0.99^{\!101}=0.9999^{\!101}<0.99^{\!-100}\times0.99^{\!101}=0.99$.

Here is an outline of a proof that $(1-x)^{\left(1-\frac{1}{x}\right)}<(1+x)^{\left(\frac{1}{x}\right)}$ when $0<|x|<1,$ 
using some of your tags such as derivatives and logarithms.
Let $f(x)=\dfrac{(1+x)^{\frac1x}}{(1-x)^{(1-\frac1x)}}=\dfrac{(1-x^2)^{\frac1x}}{1-x}.$ 
Note that $f(0)$ is not defined, but $\lim_{x\to0}f(x)=1$.
$f'(x)=-(1-x^2)^{\left(\frac1x-1\right)}\left[1+\dfrac{1+x}{x^2}\ln(1-x^2)\right].$
For $0<x<1$, $\ln(1-x^2)<-x^2<-\dfrac{x^2}{1+x}$, so $f'(x)>0.$
For $-1<x<0, \ln(1-x^2)>-\dfrac{x^2}{1+x},$ so $f'(x)<0.$
It follows that $f(x)>1$ when $0<|x|<1,$ 
and thus $(1-x)^{\left(1-\frac{1}{x}\right)}<(1+x)^{\left(\frac{1}{x}\right)}$ when $0<|x|<1.$
