Application of Euler's totient function to find last digits Q:what are the last five digits of the number $2018^{2017^{.^{.^{.^{2^{1}}}}}}$.
My Approach:I know how to find the last two digits of $N=2018^{2017^{k}} $ by$N=2018^{2017^{k}\pmod{\phi(25)}}\pmod {25}\equiv 2018^{2017^{2016^b\pmod{8}}\pmod{20}}\pmod {25}\equiv 2018^{2017^0\pmod{20}}\pmod {25}\equiv 2018^1\pmod {25}\equiv18\pmod{25}$Again,$N=18+25n\equiv0\pmod4$ Suppose n is an arbitrary integer.Solving it to find n.$18+25n\equiv0\pmod4\Rightarrow n=2\pmod4 $ Therefore, we have:$N=25(2)+18\pmod{100}\equiv68\pmod{100}$ By chinese remainder theorem.
but now  I couldn't think of how to go with it by $\pmod{100000}$.Any hints or solution will be appreciated.Thanks in advance.

Let $\psi(0)=1$ and $\psi (n)=n^{\psi (n-1)} $.
We have to find $\psi (2018)\pmod {10^5} $.
Clearly $\psi (2015)\equiv 0\pmod {5^2} $ and $\psi (2015)\equiv 1\pmod 4$, from which $\psi (2015)\equiv 25\pmod {4\cdot 5^2} $, by chinese remainder theorem.
Consequently, $\psi (2016)\equiv 2016^{25}\equiv 16^{25}\equiv 2^{100}\equiv 1\pmod {5^3}$.
On the other hand, $\psi (2016)\equiv 0\pmod 4$, from which $\psi (2016)\equiv 376\pmod {4\cdot 5^3} $, by chinese remainder theorem.
Consequently, $\psi (2017)\equiv 2017^{376}\equiv 142^{376}\equiv 406\pmod {5^4} $.
On the other hand $\psi (2017)\equiv 1\pmod 4$, from which $\psi(2017)\equiv 2281\pmod {4\cdot 5^4} $, by chinese remainder theorem.
Consequently, $\psi (2018)\equiv 2018^{2281}\equiv 11768\pmod {5^5} $.
On the other hand, $\psi (2018)\equiv 0\pmod {2^5} $, from which $\psi (2018)\equiv 36768\pmod {10^5} $, by chinese remainder theorem.
