It is well known that the limit of the function $1-(1-\frac{1}{x})^x$ is $(1-1/e)$ when $x$ goes to infinity. In addition, I know that the function is always above its limit, i.e., for every $x \geq 1$ it holds that $1-(1-\frac{1}{x})^x \geq (1-1/e)$. However, I do not know how much we can reduce from the function and still make it above the limit; hence, the following question.
What is the maximal constant $c>0$ such that there is another constant $d>1$ such that the function $f(x) = 1-(1-\frac{1}{x})^x -\frac{1}{x^c} -(1-1/e)$ satisfies $f(x) < 0$ for every $x>d$.
Here is an entirely unoriginal derivation of the initial asymptotics of your expression.
You can get explicit coefficients for the big-oh terms, but I am too lazy to do that here.
$\begin{array}\\ (1-1/x)^x &=\exp(x\ln(1-1/x))\\ &=\exp(-x(1/x+1/(2x^2)+O(1/x^3)))\\ &=\exp(-1-1/(2x)+O(1/x^2))\\ &=\dfrac1{e}(\exp(-1/(2x)+O(1/x^2))\\ &=\dfrac1{e}(1-1/(2x)+O(1/x^2))\\ &=\dfrac1{e}-\dfrac1{2ex}+O(1/x^2)\\ \end{array} $
This shows why the terms are below the limit.