Hensel's Lemma and direct formula: Why can one set a single $\overline{f'(a)}$ for all j's in $a_{j+1} = a_j - f(a_j)\overline{f'(a)}$

Question

In lecture 7 of MIT's Theory of Numbers there is a method of (using Hensel's Lemma) finding solutions for $f(x) \equiv 0 \pmod{p^{j+1}}$, where $f(x) \in \mathbb{Z}[x], f(a) \not\equiv 0 \pmod{p^j}$ and $f'(a) \not\equiv 0 \pmod{p}$. Let $a_j$ be a root of $f(x) \pmod{p^j}$, then $a_{j+1}=a_j - f(a_j)\overline{f'(a)} \pmod{p^{j+1}}$ is a root of $f(x) \pmod{p^{j+1}}$. It is written that one can choose for all j's single $\overline{f'(a)}$ such that $f'(a)\overline{f'(a)} \equiv 1 \pmod{p}$. Then $f(a_j) \equiv 0 \pmod{p^j}$ for $j \ge 1$ as long as $f'(a) \not\equiv 0 \pmod{p}$. Why can one choose only one $\overline{f'(a)}$ for, say $j = 1$, and use it for all j's? Or why does $f(a_j) \equiv 0 \pmod{p^j}$ for $j \ge 1$ hold?