Question on an example in Milne's ANT about factorisation & ramification

Question

At the end of pg 65 of Milne's ANT notes (https://www.jmilne.org/math/CourseNotes/ANT.pdf), he claimed $X^4+X^3+X^2+X+1\equiv (X+4)^4$ (mod 5) to be obvious. I understand that LHS is Eisenstein at 5 by substituting $X=Y-1$, and that if $f$ is Eisenstein at $p$, then $p$ ramifies totally in $\mathbb{Q}[\alpha]$ for $\alpha$ root of $f$. However this proposition appears in the section after this example. So I think there is another way of seeing this equality quickly? Including the factor should be $x+4$? I would imagine he treated $x+4$ as $x-1$ and try to connect with roots of unity.

Answer 1

$$\frac{X^5-1}{X-1}\equiv_5 \frac{(X-1)^5}{X-1}$$

The equality $(X-1)^5\equiv_5 X^5+(-1)^5\equiv_5 X^5-1$ follows from the binomial theorem:

$$(a+b)^p=\sum_{i=0}^{p}\binom{p}{i}a^ib^{p-i}\equiv_p a^p+b^p$$ since $\binom{p}{i}\equiv_p 0$ for $i\neq 0, p$