proof for $n(1-\frac{k+1}{n})^{n\ln(k+1)/(k+1)}<ne^{-\ln(k+1)}$

Question

I came across this inequality in a graph theory book, couldn't figure how to prove it.

$$n\left(1-\frac{k+1}{n}\right)^{n\ln(k+1)/(k+1)}<ne^{-\ln(k+1)}.$$

$n$ and $k$ are both positive integers. (Amount of vertices and minimum degree if that matters.)

Answer 1

Starting from the well know inequality $\log(1+x)<x$ for $x\neq 0$ (proof, proof), we get \begin{align} \ln\left(1-\frac{k+1}{n}\right)&<-\frac{k+1}n\\ \frac n{k+1}\ln\left(1-\frac{k+1}{n}\right)&<-1\\ \frac{n\ln(k+1)}{k+1}\ln\left(1-\frac{k+1}{n}\right)&<-\ln(k+1)\\ \ln\left(1-\frac{k+1}{n}\right)^{n\ln(k+1)/(k+1)}&<-\ln(k+1)\\ \left(1-\frac{k+1}{n}\right)^{n\ln(k+1)/(k+1)}&<e^{-\ln(k+1)}\\ n\left(1-\frac{k+1}{n}\right)^{n\ln(k+1)/(k+1)}&<ne^{-\ln(k+1)}\\ \end{align}