Why is $(1 - \frac{1}{n^{1-\epsilon}})^{n} < e^{-n^{\epsilon}}$ for $0 < \epsilon < 1$?

Question

This argument appears in one proof in my lecture and I don't know why this holds. Maybe someone knows a theorem that implies this inequality? Thanks for help.

Answer 1

If we take $x=e^{-n^\varepsilon}$ then the inequality to be proved is $$\left(1-\frac xn\right)^n<e^{-x}\tag1$$ where $x=n^{-\varepsilon}$. Taking logarithms, (1) is equivalent to $$\ln\left(1-\frac xn\right)<-\frac xn\tag2$$ at least as long as $n>x$. This is the same as $$\ln(1-y)<-y\tag3$$ for $y=x/n$. But (3) is valid for $0<y<1$; draw the graph and its tangent at $y=0$.