The sequence $(a_n)$ is defined by $a_1 = 1,$ $a_2 = 2,$ and
\[a_n^2 - a_{n - 1} a_{n + 1} = 1\]for all $n \ge 2.$  Find $a_{100}.$
This problem involves a sequence of numbers that are related by a quadratic equation. To find $a_{100},$ I need to find a way to express the general term $a_n$ in terms of $n.$ One possible approach is to try to find a pattern or a formula for the sequence by looking at the first few terms. I can calculate the first few terms by plugging in $n = 2, 3, 4, 5$ into the given equation. For $n = 2,$ I get $a_2^2 - a_1 a_3 = 1,$ which implies that $a_3 = \frac{a_2^2 - 1}{a_1} = \frac{2^2 - 1}{1} = 3.$ For $n = 3,$ I get $a_3^2 - a_2 a_4 = 1,$ which implies that $a_4 = \frac{a_3^2 - 1}{a_2} = \frac{3^2 - 1}{2} = 4.$ For $n = 4,$ I get $a_4^2 - a_3 a_5 = 1,$ which implies that $a_5 = \frac{a_4^2 - 1}{a_3} = \frac{4^2 - 1}{3} = 5.$ For $n = 5,$ I get $a_5^2 - a_4 a_6 = 1,$ which implies that $a_6 = \frac{a_5^2 - 1}{a_4} = \frac{5^2 - 1}{4} = 6.$ I notice that the sequence is just the natural numbers from $1$ to $6,$ and I wonder if this pattern continues for all $n.$ If so, then the general term $a_n$ would be $n,$ and $a_{100}$ would be $100.$ To check this, I can try to prove that $a_n = n$ for all $n$ by induction. I have already shown that $a_1 = 1$ and $a_2 = 2.$ Now, I assume that $a_n = n$ for some $n \ge 2,$ and I try to show that $a_{n + 1} = n + 1.$ By the given equation, I have $a_n^2 - a_{n - 1} a_{n + 1} = 1.$ Substituting $a_n = n$ and $a_{n - 1} = n - 1,$ I get $n^2 - (n - 1) a_{n + 1} = 1.$ Solving for $a_{n + 1},$ I get $a_{n + 1} = \frac{n^2 - 1}{n - 1}.$ I can simplify this expression by factoring the numerator and canceling out a factor of $n - 1.$ I get $a_{n + 1} = \frac{(n + 1)(n - 1)}{n - 1} = n + 1.$ This completes the inductive step, and I have shown that $a_n = n$ for all $n \ge 1.$ 