Limit of $(1-2^{-x})^x$ I am observing that $(1-2^{-x})^x \to 1$ as $x \to \infty$, but am having trouble proving this. Why does the $-x$ "beat" the $x$?
I thought of maybe considering that $$1-(1-2^{-n})^n = 2^{-n}(1+2^{-n}+2^{-2n}+\cdots+2^{-n(n-1)})\le2^{-n} \frac{1}{1-2^{-n}} \to 0,$$
as $n \to \infty$ and then noting that $(1-2^{-x})^x$ is continuous to go from integers $n$ to any real $x$.
Is there a more elegant solution, or any better intuition? Moreover, it seems that $(1-2^{-ax})^{bx}\to 1$ holds as well for any $a,b>0$.

Look at the logarithm:
$$
\log(1-2^{-x})^x=x\log(1-2^{-x})\sim x\cdot(-2^{-x})\rightarrow 0.
$$
So
$$
(1-2^{-x})^x \rightarrow 1.
$$
