Why is $\left(1-\frac{1}{k}\right)^t < e^{-t/k}$?

Question

I came across this statement, but can't see why it holds: $\left(1-\frac{1}{k}\right)^t < e^{-t/k}$

I'm sure it's something simple, but I don't have a great deal of mathematical experience. I have tried using $e^k = \sum_{n=0}^\infty \frac{\lambda^n}{n!}$, but without success.

Help would be appreciated. Thanks!

Answer 1

Using the second derivative we see that the function $x\mapsto \log(1+x)$ is concave on the interval $(-1,+\infty)$ hence its curve is below the tangent line at $x=0$ with equation $y=x$ and then we deduce the inequality $$\log(1+x)< x,\quad\forall x>-1, \;x\ne0$$ so taking $x=-\frac1k$ we find $$\log\left(1-\frac1k\right)<-\frac1k,\quad\forall k\ge1$$ then multiplying by $t$ and taking the exponential function gives the desired result.