Prove that if
$b^d$ $=$ $r \pmod n$
$b^{d/q}$ $=$ $x \pmod n$,
then
$x^q$ $=$ $r \pmod n$
for any integers $b$, $n$, $r$, and $q$ (which divides $d$).
Or more simply that $b^d$ $=$ $x^q$ $\pmod n$ if the first two conditions above are true.
Prove that if
$b^d$ $=$ $r \pmod n$
$b^{d/q}$ $=$ $x \pmod n$,
then
$x^q$ $=$ $r \pmod n$
for any integers $b$, $n$, $r$, and $q$ (which divides $d$).
Or more simply that $b^d$ $=$ $x^q$ $\pmod n$ if the first two conditions above are true.
By the Congruence Power Rule we can raise the $2$nd congruence to power $\,q,\,$ i.e.
$$\begin{align} (x\, &\equiv\, b^{\large \,d/q})^{\large \,q}\\[0.1em] \Rightarrow\,\ x^{\large q} &\equiv\, b^{\large d} \equiv\, r \end{align}\quad $$