How to prove $e^{-x}-\frac{1}{n}< \left (1-\frac{x}{n} \right )^{n}$?

Question

I was looking the solution given by Simply Beautiful Art of the following problem: https://math.stackexchange.com/a/2029616/952348 In a part of the solution the author claims that

How I can prove the left side of the inequality?, I have tried using the taylor expansion of the exponential function but I did not get anything.

Answer 1

Denote $x=nu$ and consider for $0\leq u\leq 1$ the following function $$f(u)=((1-u)e^u)^n+\frac{e^{nu}}{n}-1.$$ Then $f(0)=1/n>0$ and{*} $$f'(u)=e^{nu}(-nu(1-u)^{n-1}+1)\geq 0$$ which implies that $f$ is positive on $(0,1),$ a fact equivalent to the desired identity.

Edit: * thanks to user48203 who points out a mistake and observe

$$nu(1-u)^{n-1}\leq \sum_{k=0}^nC^k_nu^k(1-u)^{n-k}=1$$

Answer 2

Let $f(x)=(1-\tfrac xn)^n-e^{-x}+\tfrac1n.$ Then $f'(x)=-(1-\tfrac xn)^{n-1}+e^{-x}$ and if $f(x^*)$ is local minumum in $(0,n)$, then $f(x^*)=\tfrac1n-\tfrac{x^*}n(1-\tfrac{x^*}{n})^{n-1}$.

Let $g(t)=\tfrac1n-\tfrac{t}n(1-\tfrac{t}{n})^{n-1}$. Then local minumum of $g(t)$ in $(0,n)$ is $g(1)=\frac1n-\frac1n(1-\tfrac1n)^{n-1}>0.$ Hence, $f(x^*)>g(1)>0$ and $f(x)>0$ in $(0,n)$.