Let $x_1 = 9$ and the n-th term of the sequence is generated by this rule $\ x_n = 9^{x_{n - 1} } \ $.
Task : Find the last two digits of the number $\ x_{2013} \ $. I already solved the problem, but I'm interested in a more elegant take on the matter at hand. Any ideas ?
Let $a=x_{2012}$ and $b=x_{2013}$. We have $b=(-1+10)^a$. By the Binomial Theorem $b$ is congruent to $-1+10a$ modulo $100$. But $a$ is congruent to $9$ modulo $10$. It follows that $b$ is congruent to $89$ modulo $100$.
Remark: One could excise congruence notation. The last two digits of $b$ are the same as the last two digits of $-1+10a$. But $a$ has last digit $9$, so is of the shape $10k+9$. It follows that the last two digits of $b$ are the same as the last two digits of $-1+10(10k+9)$,