Exercise 11 of section 3.2.2(The art of computer programming)

Question

I'm asking about its part a.

a) If $f(z),\space a(z),\space b(z)$ are polynomials with integer coefficients, let us write $a(z)\equiv b(z) (\operatorname{mod} f(z)\space and\space m)$ if $a(z) = b(z) + f(z)u(z) + mv(z)$. Prove that the following statement holds when $p^e>2$ (p prime) , $f(0)=1$:

If $z^\lambda \equiv 1 (\operatorname{mod} f(z)\space and\space p^{e})$ and $z^\lambda \not\equiv 1 (\operatorname{mod} f(z)\space and\space p^{e+1})$ then $z^{p\lambda} \equiv 1 (\operatorname{mod} f(z)\space and\space p^{e+1})\space but\space z^{p\lambda} \not\equiv 1 (\operatorname{mod} f(z)\space and\space p^{e+2})$

Here's my reasoning:
We have $z^\lambda$ has the form of $$f(z)u(z) + p^ev(z) + 1$$ where $p^{e+1}|f(z)u(z),f(z)|p^ev(z)$ and $v(z)\not\equiv 0(\operatorname{mod} p)$ (*)
Or we can just simplify it as: $z^\lambda = Z + 1$, where Z is a multiple of $f(z)$ and $p^e$ but not $p^{e+1}$'s one. So:$$z^{p\lambda} = 1+{p\choose 1}Z+{p\choose 2}Z^2+...$$
$z^{p\lambda}-1$ is certainly divisable by f(z). Also, we can see that ${p\choose k}Z^k(1\leqslant k\leqslant p-1)$ is a multiple of $p^{e+1}$ and $p^{e+2}|Z^p$(due to the fact $p^e>2$,$p^{pe}|Z^p$ implies $p^{e+2}|Z^p$). We now can conclude: $$z^{p\lambda} \equiv 1 (\operatorname{mod} f(z)\space and\space p^{e+1})$$

Next, it's easy to establish: $p^{e+2}|p^{(e+1)k}$,where $k\geqslant$ 2. And that shows: $$z^{p\lambda} \equiv 1 + p^{e+1}v(z) (\operatorname{mod} p^{e+2}\space and\space f(z))$$ Due to (*),$1+ p^{e+1}v(z) \not\equiv 1(\operatorname{mod} p^{e+2})$, completing the proof

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My proof is based on the answer of excercise 11. And the difference between it and mine makes me confused.

If $p^{e+1}v(z) \equiv 0 (\operatorname{mod} f(z)\space and\space p^{e+2})$, there must exist $a(z)$ and $b(z)$ such that $p^{e+1}(v(z) +pa(z)) = f(z)b(z)$. Since $f(0)=1$, this implies $p^{e+1}|b(z)$(by Guass's lemma 4.6.1G); hence $v(z) \equiv 0 (\operatorname{mod} f(z)\space and\space p)$, a contradiction.

So, my questions are:
1. Why is the above argument necessary?
2. Is the Gauss's lemma 4.6.1G the one in wikipedia?