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The Fréchet distribution: $$\Phi_\alpha(x)=\begin{cases}0,\; x\leq 0\\e^{-x^{-\alpha}},\; x>0\end{cases}$$
shows a power law decay at the tail (survival):
$$1- \Phi_\alpha(x) = 1 -e^{-x^{-\alpha}}\sim x^{-\alpha}, \; x \to \infty$$
How can I prove this? I suppose it is done with a Taylor series (?)
By definition, $f$ and $g$ are asymptotically equal (as $x$ approaches infinity), denoted by
$$ f(x) \sim g (x) \; \; (x\rightarrow \infty),$$ if
$$\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)} = 1. $$
In your case, we have ($\alpha >0$)
$$\lim_{x\rightarrow \infty} \frac{1-e^{-x^{-\alpha}}}{x^{-\alpha}} \stackrel{y=x^{-\alpha}}{=}\lim_{y\rightarrow 0} \frac{1-e^{-y}}{y} \stackrel{L'Hospital}{=} \lim_{y\rightarrow 0} \frac{e^{-y}}{1} = 1 $$
