A finite group of prime order is cyclic and every nonidentity element generates it

Given: A finite group $G$ with identity $e$, with $\iota(|G|)$ prime, and an element $g\in G$ with $g\ne e$.

A prime integer $p$ satisfies $p>1$, and every positive divisor of $p$ is $1$ or $p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$).

The natural $\operatorname{ord}(g)$ is positive, equals $1$ exactly when $g=e$, and its image in $\mathbb Z$ divides $\iota(|G|)$; the embedding $\iota:\mathbb N\to\mathbb Z$ is injective and preserves order (The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity, The order of every element of a finite group divides the order of the group, The naturals embed in the integers).

The cyclic subgroup $\langle g\rangle$ is a subgroup with $|\langle g\rangle|=\operatorname{ord}(g)$ (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$, and cyclic groups, $\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$, and every cyclic group is abelian, If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$, the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$).

If $A\subseteq B$ are finite and $|A|=|B|$, then $A=B$ (A subset of a finite set is finite, with $\lvert B\rvert \le \lvert A\rvert$, and equality holds if and only if $B = A$).

If a finite set $G$ contains $e$ and $|G|\ne1$, then some element of $G$ differs from $e$: otherwise $G=\{e\}$, whose cardinality is $1$ (The cardinality $\lvert A\rvert$ of a finite set).