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Isolating $a_{n+1},$ we find $$a_{n+1}=\frac{a_n^2-1}{a_{n-1}}.$$Then $$\begin{array}{rl}{a}_3& =\frac{a_2^2-1}{a_1}=\frac{2^2-1}{1}=3,\\ a_4& =\frac{a_3^2-1}{a_2}=\frac{3^2-1}{2}=4,\\ a_5& =\frac{a_4^2-1}{a_3}=\frac{4^2-1}{3}=5,\end{array}$$and so on.

By induction, we can show that $a_n=n$ for all positive integers $n.$ In particular, $a_{100}=\boxed{100}.$