Why is $(1-\frac{1}{n})^{n-k} \geq e^{-1}$? This inequality is written on page 2 of https://www.cs.cmu.edu/~avrim/Randalgs11/lectures/lect0202.pdf.
As an extension, can we say anything about relating $(1-\frac{j}{n})^{n-k}$ and $(\frac{j}{n})^k$ to an exponentiation, for any $j$?
I presume $n$ and $k$ are positive integers. Then $(1-1/n)^{n-k}\ge(1-1/n)^{n-1}$ so let's assume $k=1$. Then $$\ln\left(1-\frac1n\right)^{n-1}=-(n-1)\left(\frac1n+\frac1{2n^2} +\frac1{3n^3}+\cdots\right)=-\left(1-\frac1{2n}-\frac1{6n^2}-\cdots\right).$$