China Girls' Mathematical Olympiad

We have $a_{n+1}-1=a_n(a_n-1)$, and then $$\frac{1}{a_{n+1}-1}=\frac{1}{a_n-1}-\frac{1}{a_n}.$$ So, $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_{2^{003}}}=(\frac{1}{a_1-1}-\frac{1}{a_2-1})+(\frac{1}{a_2-1}-\frac{1}{a_3-1})+\cdots +(\frac{1}{a_{2^{003}}-1}-\frac{1}{a_{2^{004}}-1})$$ $$=\frac{1}{a_1-1}-\frac{1}{a_{2^{004}}-1}=1-\frac{1}{a_{2^{004}}-1}.$$

It is easy to see that $\{a_n\}$ is strictly monotonic increasing, so $a_{2^{004}}>1$, and that means $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_{2^{003}}}<1.$$ In order to prove the left inequality, we only need to prove that $a_{2^{004}}-1>2^{003^{2^{003}}}$. By induction, we have $a_{n+1}=a_n{a}_{n-1}\cdots a_1+1$ and $a_n{a}_{n-1}\cdots a_1>n^n$ ($n\ge 1$). This completes the proof.