Proof of Hensel's Lemma not clear

Question

If you look at the following proof of Hensel's Lemma http://isites.harvard.edu/fs/docs/icb.topic1472247.files/Hensels%20lemma.pdf you will see that the author determines the conditions which these elements $x_n$ must satisfy, then solves for what $x_n$ has to be as a result. They do not actually show that the $x_n$ have the given properties that they claim. In particular, they make the claim that $$w+f'(x_n)S \equiv 0 \pmod{\theta}$$ But if you look at how $w$ and $S$ are defined in the first place, it is not at all clear that this should be the case.

Answer 1

The claimed congruence $w+f'(x_n)S\equiv 0 \bmod \theta$ is clear from the line above, if you just divide by $\theta^n$, writing $f(x_n)=w\theta^n$: the congruence modulo $\theta^{n+1}$ is derived from the Taylor expansion $$ f(x_n+\theta^nS)=f(x_n)+f'(x_n)\theta^nS+\frac{1}{2!}f''(x_n)\theta^{2n}S^2+\cdots + $$ Now consider this modulo $\theta^{n+1}$. Then $$ w\theta^n+f'(x_n)\theta^n S \equiv 0\bmod \theta^{n+1}, $$ hence $$ w+f'(x_n)S\equiv 0 \bmod \theta. $$