= Solution
For a direct check of <unramified> behavior, the <cyclotomic polynomial> $\Phi_{25}$ divides $X^{25}-1$. Modulo $7$, the latter has derivative $25X^{24}=4X^{24}$ and has no common root with that derivative. Thus $\overline\Phi_{25}$ is separable, so $7$ does not divide its <polynomial discriminant>. The <field discriminant> divides this polynomial discriminant, and therefore $7$ is <unramified> in $\mathbb Q(\zeta_{25})$. The <Frobenius automorphism> satisfies
$$
\boxed{\operatorname{Frob}_7(\zeta_{25})=\zeta_{25}^{\,7}.}
$$
Indeed its residue action is seventh powering. The reduced $25$th roots of unity are distinct because $7\nmid25$, so the residue congruence identifies the exponent uniquely modulo $25$.
The <Galois group> is $(\mathbb Z/25\mathbb Z)^\times$, of order $\varphi(25)=20$. Since $7^2=49\equiv-1\pmod{25}$, $7^4\equiv1$ but $7^2\not\equiv1$. Consequently the <Frobenius automorphism> has order four. An <unramified> <decomposition group> is generated by that element, and the <orbit-stabilizer theorem> gives
$$
\boxed{g=20/4=5,\qquad f=4,\qquad e=1.}
$$
There are therefore \b[five primes above $7$], each with <residue field> $\mathbb F_{7^4}$.
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