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$t\equiv \frac{f(r_{k-1})\overline{f'(r)}}{p^{k-1}}\mod p$, not $\frac{f(r)\overline{f'(r)}}{p^{k-1}}\mod p$ with the explanation underlined. I have included an example of the lifting proof for reference: enter image description here

Why is $f(r_{k-1})\not\equiv f(r)$ in the formula $r_k=r_{k-1}-f(r_{k-1})\overline{f'(r)}$ just like $f'(r_{k-1})\equiv f'(r_1)$? Originally the formula is $r_k=r_{k-1}-f(r_{k-1})\overline{f'(r_{k-1})}$

Please note that the formula I made up $r_k=r_{k-1}-f(r)\overline{f'(r)}$ does not work on any of the examples I tried...

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    Not sure I understand your question. The purpose of this corollary is to show that you can take a solution $r_k$ for $f(x) \equiv 0 \mod{p^k}$ and generate a "better" solution $r_{k+1}$ for $f(x) \equiv 0 \mod{p^{k+1}}$. The initial solution is $r=r_1$. – 2017-01-25
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    Still not clear. The proof looks correct to me. Maybe you can write down an example you tried that doesn't seem to work. – 2017-01-25
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    The proof is correct. I'm wondering why if $r_{k-1}\equiv r \pmod p$ why isn't $f(r_{k-1}) \equiv f(r_1) \mod p$ in the formula $t\equiv \frac{f(r_{k-1})\overline{f'(r)}}{p^{k-1}}\mod p$. You have to understand Hensel's lemma to understand the corollary. Originally f'(r) was f'(r_{k-1}). Should I add that proof in to make my question make sense? – 2017-01-25
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    $f(r_{k-1})$ is congruent to $f(r_1)$ mod $p$. But why do you want to do that? $r_1$ is only a solution to the congruence mod $p$, so it is a "worse" solution than $r_{k-1}$, which is a solution mod $p^{k-1}$. – 2017-01-25
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    Well, it was just an idea so that you didn't have to calculate intermediate steps to reach $r_k$ from $r_1$ but the formula would still need $r_{k-1} \dots$ unless you could convert $r_{k-1}$ into $r_1$ in closed form which is circular because that closed form is what I was thinking about. But why doesn't it work if you solve for $\require{enclose} t\equiv \frac{\enclose{horizontalstrike}{f(r_{k-1})}f(r_{1})\overline{f'(r)}}{p^{k-1}}\mod p$ and then plug it into $r_{k-1} -t*p_{k-1} \mod p_k$, t is $\mod p$ for every k. – 2017-01-25

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