I know it can be done when $xn \ll 1 $, but what about the cases when $xn \gt 1$ ?
My best try is to use sth like: \begin{align*} (1-x)^n &= \sum\limits_{j=0}^{\infty}\left( \begin{array}{c} n \\ j \end{array} \right)(-x)^j \end{align*}
However, when $n$ is very large, calculation of the binomial coefficients may exceed computational capability of available tools such as mathwork, so I am stuck.
On
I was also interested in the same question. Here is an answer that uses ideas from the answer and comment above. Since $\log(1-x)^n = -n\sum_{j=1}^\infty x^j/j$, then we have $$\begin{align}(1-x)^n &= \exp\left(-n\sum_{j=1}^{\infty} \frac{x^j}{j}\right) \\&=\exp(-nx).\exp\left(-n\sum_{j=2}^{\infty}\frac{x^j}{j}\right) \end{align}$$ Now let's find the absolute error of approximating $(1-x)^n$ with $e^{-nx}$. Subtracting $e^{-nx}$ from both sides and taking absolute values we get $$\begin{align}|(1-x)^n-e^{-nx}| &= |e^{-nx}(e^{-n\sum_{j=2}^\infty x^j/j}-1)| \\&= e^{-nx}(1-e^{-n\sum_{j=2}^\infty x^j/j}) \\&<e^{-nx} \end{align} $$ which tells us that the error in this approximation is bounded by $e^{-nx}$ (which goes to 0 as $nx\to \infty$).
For $0<x<1$ we have $\log (1-x)^n=-n\sum_{j=1}^{\infty}x^j/j$. Taking $k$ large enough that $n\sum_{j>k}x^j/j<r$, and let $\sum_{j=1}^{k}x^j/j=F(x,k). $ We have $e^ {-n F(x,k)}e^{-r}<(1-x)^n<e^{-n F(x,k)}.$ Of course $k$ depends on both $n$ and $x$ for any given $r>0$, but given $y\in (0,x)$ we can use the same $k$.