solving $\left( m'\right) ^{d}\equiv m\cdot m^{r\left( p-1\right) \left( q-1\right) }\left( mod\ p\right) $

Question

maybe someone can help: I am trying to follow a lecture and there is:

given : $\left( m'\right) ^{d}\equiv m\cdot m^{r\left( p-1\right) \left( q-1\right) }\left( mod\ p\right) $

and : $ m^{p-1} \equiv 1 \ \left(mod\ p\right)$

then : $\left( m'\right) ^{d}\equiv m \ \left( mod\ p\right) $

I don't understand why the whole term $ m^{r\left( p-1\right) \left( q-1\right)}$ is cancelled.

Your help is appreciated!!

Answer 1

$m^{p-1}\equiv 1(\mod p)$ implies $m^{r(p-1)(q-1)}=(m^{p-1})^{r(q-1)} \equiv 1^{r(q-1)}(\mod p) =1$.