Proving $2$ generates $(\mathbb{Z}/19\mathbb{Z})^*$ by only looking at a few powers

If $2$ does not generate the whole order-18 group, it generates a proper subgroup of order dividing $18$, that is, of order $1$ or $2$ or $3$ or $6$ or $9$, which means that one of $2^1, 2^2, 2^3, 2^6, 2^9$ would be $\equiv 1$.

Um, actually ... it suffices to know that $2^9\not\equiv 1$ and $2^6\not\equiv 1$. Do you see why?

By Fermat $\,2^{\large 18} \equiv 1,\,$ thus $\, 2\,$ has order $18\,$ iff $\,2^{\large 6}\!\not\equiv 1$ and $\,2^{\large 9}\!\not\equiv 1\,$ by the following

Order Test $\ \,a\,$ has order $\,n\iff a^{\large n} \equiv 1\,$ but $\,a^{\large n/p} \not\equiv 1\,$ for every prime $\,p\mid n.\,$

Proof $\ (\Leftarrow)\ $ If $\,a\,$ has $\,\color{#c00}{{\rm order}\ k}\,$ then $\,k\mid n\,$ [proof]. If $\:k < n\,$ then $\,k\,$ is proper divisor of $\,n\,$ therefore $\,k\,$ arises by deleting at least one prime $\,p\,$ from the (unique) prime factorization of $\,n,\,$ hence $\,k\mid n/p,\,$ say $\, kj = n/p,\ $ so $\ a^{\large n/p} \equiv (\color{#c00}{a^{\large k}})^{\large j}\equiv \color{#c00}1^{\large j}\equiv 1,\,$ contra hypothesis. $\ (\Rightarrow)\ $ Clear.