= Solution
The <characteristic of a field>[characteristic] is the smallest positive integer $p$ with $p1_F=0$. Such an integer exists because the additive <group> is finite. If it factored as $p=ab$ with $1<a,b<p$, then $(a1_F)(b1_F)=0$ with both factors nonzero, contradicting the absence of zero divisors. Hence it is <prime>. The elements generated by $1_F$ form a subfield isomorphic to $\mathbb F_p$. The finite set $F$ is a <finite-dimensional vector space> of some <dimension> $a\ge1$ over that subfield, so $\boxed{|F|=p^a}$.
A nonzero <polynomial> of <polynomial degree> $n$ over a <field> has at most $n$ distinct roots. This follows by <induction>: a root $r$ gives a factor $X-r$, and every different root must be a root of the remaining factor by cancellation. Thus $X^n-1$ has at most $n$ roots. In the finite multiplicative <group>, the elements whose order divides $n$ are exactly those satisfying $x^n=1$. The supplied cyclicity criterion therefore proves that \b[the multiplicative <group> is cyclic].
For the <prime> of the specified form write $m=2^r s$ with $s$ odd. If $s>1$, then $2^{2^r}+1$ is a proper factor of $(2^{2^r})^s+1$, impossible. Thus $m$ is a <power of two>. Since $2^m\equiv-1\pmod p$, the order of two divides $2m$ but does not divide $m$. All divisors of the <power of two> $2m$ other than $2m$ itself divide $m$. Consequently
$$
\boxed{\operatorname{ord}_p(2)=2m.}
$$
It generates the multiplicative <group> precisely when $2m=p-1=2^m$. For positive integers $m$ this holds exactly at $m=1,2$: from $m=3$ onward $2^m>2m$, and the inequality persists on increasing $m$. These give $\boxed{p=3\text{ or }5}$, equivalently the stated range among Fermat <primes>.
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