Let's say that $p$ is prime number, $z$ is natural number and our form $\frac{p-1}{2^z}=n$ and is even. $p>n$. From binomial theroem $n^x$ ($x$ is natural number) = form in photo. So is $$n^x \equiv \bigg(\frac{1}{2^z}\bigg)^x \pmod{p}$$ the truth ? Better to look at photo so you can properly see what I want to show you.
I don't know why, beacuse it makes kinda sense to me, it is not true. For example $p=13$ and $z=1$ and $x=p-2$ or any other I think. Also I don't know if mod can be written as fraction and if it is written like that what I need to do to make it natural number?
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That is not necessarily true. $$\bigg(\frac{p-1}{2^z}\bigg)^x\equiv\bigg(\frac{-1}{2^z}\bigg)^x\pmod{p}$$
So only if $x$ is even you have $$\bigg(\frac{p-1}{2^z}\bigg)^x\equiv\bigg(\frac{1}{2^z}\bigg)^x\pmod{p}$$
EDIT: you cannot always find the residue. $\frac{1}{2^z}$ is the modular inverse of $2^z$, ususally denoted with $2^{-z}$ Those do not have a formula. Read more here.