1. [5] Define the sequence of positive integers $a_{n}$ recursively by $a_{1}=7$ and $a_{n}=7^{a_{n-1}}$ for all $n \geq 2$. Determine the last two digits of $a_{2007}$.
Answer: 43. Note that the last two digits of $7^{4}$ are 01. Also, $a_{2006}=7^{a_{2005}}=(-1)^{a_{2005}}=-1=3$ $(\bmod 4)$ since $a_{2005}$ is odd. Therefore, $a_{2007}=7^{a_{2006}}=7^{3}=43(\bmod 100)$.