Singapore Mathematical Olympiad (SMO)

We shall prove by induction that if the original numbers are $a_1, \dots, a_n$, $n \ge 2$, then the last number is $(1+a_1) \cdots (1+a_n) - 1$.
The assertion is certainly true for $n=2$, the base case. Now suppose it is true for $n=k \ge 2$. Consider $k+1$ numbers $a_1, \dots, a_{k+1}$ written on the board. After one operation, we are left with $k$ numbers. Without loss of generality, we can assume that the student erases $a_k$ and writes $b_k = a_k+a_{k+1}+a_k a_{k+1} = (1+a_k)(1+a_{k+1})-1$. After a further $k$ operations, we are left with the number
$$(1+a_1) \cdots (1+a_{k-1})(1+b_k) - 1 = (1+a_1) \cdots (1+a_{k-1})(1+a_k)(1+a_{k+1}) - 1.$$
This completes the proof of the inductive step. Thus the last number is
$$(1 + \frac{1}{2})(1 + \frac{1}{3}) \cdots (1 + \frac{1}{2010}) - 1 = 2010$$