Fermat's little theorem:


if gcd(a, p) = 1

then

$$ a^{p-1} \equiv 1 \mod p $$


let $y^z = t$, $t = k(p-1)+b$

so $b=y^z\%(p-1)%$



$x^{y^z} \mod p$
$\equiv x^t$
$\equiv (x^{p-1})^{k} \times x^b$
$\equiv (1)^{k} \times x^b$
$\equiv x^b$
$\equiv x^{(y^z\%(p-1))}$