Isolating $ a_{n + 1}, $ we find \[a_{n + 1} = \frac{a_n^2 - 1}{a_{n - 1}}.\]Then \begin{align*} a_3 &= \frac{a_2^2 - 1}{a_1} = \frac{2^2 - 1}{1} = 3, \\ a_4 &= \frac{a_3^2 - 1}{a_2} = \frac{3^2 - 1}{2} = 4, \\ a_5 &= \frac{a_4^2 - 1}{a_3} = \frac{4^2 - 1}{3} = 5, \end{align*}and so on. By induction, we can show that $ a_n = n $ for all positive integers $ n $. In particular, $ a_{100} = \boxed{100} $.