Prove that $(1-\frac{1}{k})^d \le e^{-\frac{d}{k}} $

Question

Prove that $(1-\frac{1}{k})^d \le e^{-\frac{d}{k}} $ for $d,k \ge 0$

I know that $(1+\frac{1}{n})^n \le e$ but does that help? Actually, I don't really 'know' this, but I've heard it's true at least - soemthing about showing it's an increasing function and tends to e?

Thanks

Answer 1

Equivalent to $$\left( e^{\frac{1}{k}} \left( 1-\frac{1}{k} \right) \right)^d \leq 1.$$ Use $e^{x} \geq 1+x$ with $x=-\frac{1}{k}$.

Answer 2

A very important inequality about the exponential is $$\tag1e^x\ge 1+x\qquad \text{for all }x\in\mathbb R$$ (with equality iff $x=0$). Hence if $k\ge 1$ we can let $x=-\frac1k$ and obtain $$\tag2 1-\frac 1k\le e^{-\frac 1k}. $$ Since $1-\frac1k\ge 0$, we can take $d$th power on both sides ($d\ge1$) to obtain $$\tag3 \left(1-\frac 1k\right)^d\le e^{-\frac dk}. $$

Of course we cannot let $k=0$, but apart from that $(3)$ holds trivially for $d=0$.