Lets use only our knowledge of $e$, and a geometric series identity. Notice that $$\left(1-e^{-\frac{1}{k}}\right)\left(1+e^{-\frac{1}{k}}+e^{-\frac{2}{k}}\cdots+e^{-\frac{k-1}{k}}\right)=1-e^{-1}.$$ Now, $e^{-x}\leq1$ for $x\geq0,$ so $$\left(1-e^{-\frac{1}{k}}\right)k=\left(1-e^{-\frac{1}{k}}\right)\left(1+1+1\cdots+1\right)$$
$$\geq\left(1-e^{-\frac{1}{k}}\right)\left(1+e^{-\frac{1}{k}}+e^{-\frac{2}{k}}\cdots+e^{-\frac{k-1}{k}}\right) $$
$$=1-e^{-1}$$ $$\geq\frac{1}{e},$$ and so $$1-e^{-\frac{1}{k}}\geq\frac{1}{ke}.$$