A lower bound for $(1-e^x)^n$

Question

I want to find a lower bound for $$(1-e^x)^n$$ $n$ integer, $x$ real, and $1-e^x\geq 0$. One lower bound is (Bernoulli's inequality) $$(1-e^x)^n\geq 1-ne^x$$ But I need a tighter lower bound that is polynomial in $n$ and $x$.

I am thinking of combining binomial expansion of $(1-e^x)^n$ and Taylor series of $e^x$, but that gets too complicated. I am wondering if there exists any tight lower bound for this that I am not aware of. Any help/hint is appreciated.

Answer 1

No, there isn't.

There is no polynomial $f(n,x)$ in $n$ and $x$ such that there exist positive $c\in\Bbb R$ such that for any $x\lt -c$ and any $n\in\Bbb Z$, $$(1-e^x)^n\ge f(n,x)\geq 1-ne^x$$

For the sake of contradiction, let $f(n,x)$ be such a polynomial. Let $n=1$. We will have for x sufficient small, $1-e^x\ge f(1,x)\geq 1-e^x$. That is, $f(1,x)=1-e^x$, which cannot be true.

OK, you may say $n=1$ is an outlier. Let us exclude that outlier. However, we have a stronger statement.

Let $1\neq n\in\Bbb N$. There is no polynomial $f(x)$ such that there exist positive $c\in\Bbb R$ such that for any $x\lt -c$, $$(1-e^x)^n\ge f(x)\geq 1-ne^x$$

Here is a simple proof. For the sake of contradiction, let $f(x)$ be such a polynomial. Taking $x$ to $-\infty$, we find that $f(x)$ goes to 1. As a polynomial in $x$, $f(x)$ must be the constant polynomial $1$. Then for any $x$ and positive $n$, $(1-e^x)^n<1=f(x)$.

(So, it seems there is some typo or lapse in the question since its answer is almost trivially false)