The author (like many) is likely skipping the "degenerate" case (presumably obvious to the reader). Since you write in a comment that this is the first time you've encountered an argument like this, you may find it instructive to see how such arguments naturally generalize as below, e.g. inverting an algebraic number $\alpha$ by rationalizing the denominator of $\frac{1}{\alpha}$ (a method likely you have seen).
Generally, over any field (such as $\Bbb Z_p$) if a polynomial $f(x)$ has nonzero constant term $f(0)\neq 0$ then $x=0$ is not a root, so every root is nonzero, so invertible. Thus when doing algebra with roots of $f$ we can invert (and cancel) them. OP is the special binomial case $\,f(x) = x^n - a,\ a\!\neq\! 0$.
Generally the above shows that an $\rm\color{#c00}{algebraic}$ root of $f$ divides its simpler $\rm\color{#0a0}{integer}$ constant term, and this simpler multiple allows us to reduce $\rm\color{#c00}{algebraic}$ number inversion to $\rm\color{#0a0}{integer}$ inversion, e.g. using norms or rationalizing denominators.
You may also find instructive the case analysis in the comparison of the little Fermat Theorem in general vs. nonzero form: $\,a^p\equiv a\,$ vs. $\,a^{p-1}\equiv 1,\,$ a special case $\,f = x^{p-1}-1$ of
$$\begin{align} \forall x\!:&\ \ \ \ \ \ \ xf(x)= 0\\[.2em] \iff\ \forall x\ &\ [\:\!x\not= 0\Rightarrow f(x)= 0\:\!] \end{align}\qquad$$