Minimal polynomial of $e^{\frac{2 \pi i}{p^2}}$ (p is prime number)

Question

$X^{p^2}-1 = (X-1)(X^{p-1}+X^{p-2}+..+X+1)(X^{p(p-1)}+X^{p(p-2)}+..X^{p}+1)$

Then, $e^{\frac{2 \pi i}{p^2}}$ is a root of $f(X)=X^{p(p-1)}+X^{p(p-2)}+..X^{p}+1$

Hence, $g(X)=\frac{X^{p^2}-1}{X-1}$ for $X\neq 1$

thus, for $X \neq 0$ we have

$$g(X+1)=\frac{(X+1)^{p^2}-1}{X}=\sum_{i=1}^{p^2} \binom{p^2}{i}X^{i-1}$$

But $g(X)$ isn't irreducible and minimal polynomial

So, How can I prove it by Eisenstein Criteria?

Answer 1

The polynomial $$\frac{(x+1)^{p^2}-1}{(x+1)^p-1}$$ is Eisenstein at $p$.