jmc

Since $9\cdot 2=18=17+1$, it follows that $9$ is the modular inverse of $2$, modulo $17$. Thus, $2^n\equiv 2^9(\mathrm{mod}17)$. After computing some powers of $2$, we notice that $2^4\equiv -1(\mathrm{mod}17)$, so $2^8\equiv 1(\mathrm{mod}17)$, and $2^9\equiv 2(\mathrm{mod}17)$. Thus, $(2^9{)}^2\equiv 4(\mathrm{mod}17)$, and $(2^9{)}^2-2\equiv \boxed{2}(\mathrm{mod}17)$.

Notice that this problem implies that ${\left(a^{2^{-1}}\right)}^2\not\equiv a(\mathrm{mod}p)$ in general, so that certain properties of modular inverses do not extend to exponentiation (for that, one needs to turn to Fermat's Little Theorem or other related theorems).