Prove that if $q\mid x$ and $q^k\parallel x^p$ then $p\mid k$.

$$\begin{align} q^{\large k}\mid\mid x^{\large p}\ \Rightarrow\, \, \ q^{\large k}\ \color{#c00}n &\,=\, x^{\large p},\,\ \ \color{#c00}{q\nmid n},\, \ \text{so comparing powers of $\,q$}\\[.2em] \Rightarrow\ k+ \color{#c00}0 &\,=\, p\ \nu_q(x),\ \ \ {\rm so}\ \ \ p\,\mid\, k & \end{align}$$