= Solution
Let $a$ be a root modulo $p^j$. Every lift to a residue modulo $p^{j+1}$ is uniquely $a+tp^j$ with $t\pmod p$. The <polynomial> expansion gives
$$
f(a+tp^j)\equiv f(a)+tp^jf'(a)\pmod{p^{j+1}}.
$$
Because $a\pmod p$ is a root and there are no common roots of $f$ and $f'$, $f'(a)$ is nonzero modulo $p$. Thus precisely one $t$ solves
$$
\frac{f(a)}{p^j}+t f'(a)\equiv0\pmod p.
$$
Every root at one level therefore has exactly one lift, and every root at the next level reduces to a root at the previous level. <Induction> proves equality of the counts from $p$ to $p^l$. This is <simple-root Hensel lifting by successive residues> at finite precision; completeness is unnecessary for this finite-level assertion.
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