Prove Inequality $(1 - 1/t)^k \le \exp(-k/t)$

Question

I have seen the inequality bellow in an online lecture video. I cannot see how it holds for any $k \geq 0$.

Inequality:

$$\left(1 - \frac1t\right)^k \leq \exp\left(\frac{-k}{t}\right)$$ where $t$ is a positive constant and $k > 0$.

Is this related to some well-known property?

Answer 1

This has to do with the definition of the number $e$. $$e = \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n$$ More generally, $$e^x = \lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n$$ for any real number $x$. Moreover, the expression $(1+x/n)^n$ is increasing in $n$. Hence, $$1-\frac{1}{t} \leq \left(1+\frac{-1/t}{n}\right)^n \leq e^{-1/t}$$ Raising both sides to the $k$th power yields the result.

Answer 2

One has $1-1/t \leq e^{-1/t}$ because $e^{-x}$ is convex, so it lies above its tangent line. Raise both sides to the power $k$.