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 \pmod{17} $. After computing some powers of $ 2 $, we notice that $ 2^4 \equiv -1 \pmod{17} $, so $ 2^8 \equiv 1 \pmod{17} $, and $ 2^9 \equiv 2 \pmod{17} $. Thus, $ (2^9)^2 \equiv 4 \pmod{17} $, and $ (2^9)^2 - 2 \equiv \boxed{2} \pmod{17} $. Notice that this problem implies that $ \left(a^{2^{-1}}\right)^2 \not\equiv a \pmod{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).