= Solution
The <characteristic of a field> is the least positive <integer> $p$ with $p1_F=0$, or zero if none exists. Finiteness forces a positive characteristic, since the sequence of <integer> multiples of one must repeat. If this least <integer> factored into two smaller positive factors, their nonzero field multiples would multiply to zero, contradicting the absence of zero divisors. Hence it is prime. The field is a <finite-dimensional vector space> over its prime <subfield> $\mathbb F_p$, so choosing a <basis> gives $|F|=p^a$.
A nonzero <polynomial> of degree $n$ over a field has at most $n$ roots: each root supplies a linear factor by <polynomial> division, and induction applies to the quotient. In particular $X^n-1$ has at most $n$ roots. The elements of $F^\times$ whose orders divide $n$ are precisely those roots. The permitted finite-<group> criterion therefore proves \b[$F^\times$ is cyclic].
For the final assertion take an odd <Fermat prime> $p=2^m+1$, $m\geq1$. Primality forces $m$ to be a power of two: an odd factor $s>1$ in $m=2^rs$ would factor $(2^{2^r})^s+1$ nontrivially. Since $2^m\equiv-1\pmod p$, the order of two divides $2m$ but not $m$. As $2m$ is a power of two, its only divisor not dividing $m$ is $2m$ itself. Thus
$$
\boxed{\operatorname{ord}_p(2)=2m}.
$$
It generates the full <group> exactly when $2m=2^m$, which holds for $m=1,2$ and fails for $m\geq3$. Therefore the possibilities are \b[$p=3,5$], equivalently $p\leq5$ among odd Fermat primes. The oddness is implicit in referring to two as a unit; $p=2$ is not an admissible case for that claim.
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