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 $a,b>0$.

$(1-2^{-ax})^{bx}=e^{bx \log (1-2^{-ax})}$
We use L'Hospital's rule to calculate the limit:
$\lim_{x\to \infty}x \log (1-2^{-ax})=\lim_{x\to \infty}\frac{\log (1-2^{-ax})}{\frac{1}{x}}=\lim_{x\to \infty}\frac{-2^{-ax}\log 2 (-a)}{(1-2^{-ax})}-x^2=\lim_{x\to \infty}-a\log2 \frac{x^2}{2^{ax}-1}\to 0$ 
Hence the limit is 1 $\forall a,b>0$.
