If $(x^y)^z = x^{y\cdot z}$, why does $(-5)^{2^{0.5}}$not equal $(-5)^1$? As shown by Wolfram Alpha, $(x^2)^{0.5}$ is equal to |x|, but if you tried to simplify it to $x^{2\times {0.5}}$, it would just be $x^1$, or $x$.
Is there some unwritten rule about that distribution law that means you can't do it with fractional exponents?
Edit: What confuses me the most is how Wolfram Alpha also believes $(x^2)^{0.5}$ = x while actually showing a graph of |x|

Whenever we have grouping symbols we work from the inside out , following PEMDAS , WIA included. If you want the exponents to cancel do it this way , $$ (x^{0.5})^2$$
Your numerical example becomes $$((-5)^{0.5})^2 = ( \sqrt{-5})^2 = (i \sqrt{5})^2 = i^2 5 = -5 $$
And in general for real numbers x , $$( \sqrt{x})^2 = x$$ $$\sqrt{x^2} = |x| $$
Put it in WIA to see what it gives you.
