Find the last digits of $a_{2009}$, and of $b_{2009}$. Define the sequences $a_1, a_2,...$ and $*b_1, b_2,...*$ by $a_1 = b_1 = 7$ and 
$$a_{n+1} = {a_n}^7, \\ b_{n+1} = 7^{b_n}$$ for $n\ge 1$.
Find the last digits of $a_{2009}$, and of $b_{2009}$.
What about the last two digits or more?
I know the order of 7 mod 100 is 4. Im  not sure if that helps but how will we work this out?
 A: Part 1:$$a_1\equiv(mod10)7 $$$$a_2=7^7\equiv(mod10)(-3)^7\equiv3$$$$a_3=a_2^7\equiv(mod10)3^7\equiv7$$$$.$$$$.$$$$.$$$$a_{2009}\equiv(mod10)7$$'''''''''''''''''''''$$a_1\equiv(mod100)7$$ by a simple calculation we have $7^4\equiv(mod100)1$
$$a_2\equiv(mod100)a_1^7\equiv7^7\equiv7^3$$
$$a_3\equiv(mod100)a_2^7\equiv(7^3)^7\equiv7^{21}\equiv7(7^4)^5\equiv7$$
$$a_4\equiv(mod100)7^3$$
$$$$
and then we have:
$$a_{2009}\equiv(mod100)7
$$''''''''''''''''''''''$$$$
Part 2:$$$$
We need to show $b_n\equiv(mod4)3$$$$$
Proof by induction:$$b_1\equiv(mod4)3$$
$$b_k\equiv(mod4)3 \Longrightarrow b_{k+1}\equiv(mod4)3$$$$b_{k+1}=7^{b_k}\equiv(mod4)(-1)^{4k+3}\equiv-1\equiv3$$$$$$now we have:$$b_n\equiv(mod100)7^{b_{n-1}}\equiv7^{4k+3}\equiv7^3\equiv43$$$$$$$$b_n\equiv(mod10)7^{4k+3}\equiv7^3\equiv3$$
A: Hint: We have
$$a_n=7~\widehat~~7~\widehat~~(n-1)$$
$$b_n=7~\widehat~~ b_{n-1}=7~\widehat~~7~\widehat~~b_{n-2}=\dots$$
and in the case of finding the last two digits, the order of $7$ mod $100$ is $4$ and the order of $7$ mod $4$ is $2$, so
$$a_n\equiv7~\widehat~~(7~\widehat~~[(n-1)\bmod2]\bmod4)\pmod{100}$$
$$b_n\equiv7~\widehat~~(7~\widehat~~[b_{n-2}\bmod2]\bmod4)\pmod{100}$$
and the same strategy can be applied to find more digits.
