 Solution

The characteristic of a field is the least positive integer $p$ with $p{1}_F=0$, 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 $F^{\times }$ is cyclic.

For the final assertion take an odd Fermat prime $p=2^m+1$, $m\ge 1$. Primality forces $m$ to be a power of two: an odd factor $s>1$ in $m=2^{r}s$ would factor $(2^{2^r}{)}^s+1$ nontrivially. Since $2^m\equiv -1(\mathrm{mod}p)$, 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. ThusIt generates the full group exactly when $2m=2^m$, which holds for $m=1,2$ and fails for $m\ge 3$. Therefore the possibilities are $p=3,5$, equivalently $p\le 5$ 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.