TITLE: What is the 2017th digit (from the right) of $2017^{2016^{2015^{\cdots^1}}}$?
QUESTION [21 upvotes]: What  is  the  value of the $2017^{\rm th}$ digit starting from right side for $$ {2017^{2016^{2015^{\ldots 1 }}}}?$$ 

My attempt: $$2017^{n} \equiv  x  \pmod {10^{2017}}  \quad \Longrightarrow \quad  { 7 }^{ n } \equiv  x \pmod {{ 10 }^{ 2017 }}.$$ I stopped at this point. 
So, first of all "could you tell me some books which can improve my abilities in number theory" and finally I hope that you can help me to figure this question out.

REPLY [11 votes]: This isn't an answer just some useful information maybe, I found ${2017^{2016^{2015^{\ldots 1 }}}}\equiv 1\pmod{2^{2017}}$ and ${2017^{2016^{2015^{\ldots 1 }}}}\equiv 2017\cdot 3^{5^{2016}}\pmod{5^{2017}}$ 
Now this could probably be solved by CRT with some good calculator or if combining them you get something which has less then $2017$ digits $\mod 10^{2017}$ you can check $\log_{10}$ of that must be smaller then $2016$ then you can conclude that the $2017th$ digit is $0$.
Everything below is how I got the two modulo above.
Lets try the CRT
$$2017^{\varphi(2^{2017})}\equiv 1\pmod{2^{2017}}$$
Since $\varphi(2^{2017})=2^{2016}$ and $2^{2016}\mid 2016^{2015}$ we have that
${2017^{2016^{2015^{\ldots 1 }}}}\equiv 1\pmod{2^{2017}}$
We also have $\varphi(5^{2017})=4\cdot 5^{2016}$ $$2017^{\varphi(5^{2017})}\equiv 1\pmod{5^{2017}}$$
Now lets split ${2016^{2015^{\ldots 1 }}}\pmod{4\cdot 5^{2016}}$ by CRT
$${2016^{2015^{\ldots 1 }}}\equiv 0\pmod{4}\\{2016^{2015^{\ldots 1 }}}\pmod{5^{2016}}$$
Now lets look at
$${2015^{2014^{\ldots 1 }}}\pmod{4\cdot5^{2015}}$$
Again lets split it by CRT
$${2015^{2014^{\ldots 1 }}}\equiv (-1)^{2014^{\ldots 1 }}\equiv 1\pmod{4}\\{2015^{2014^{\ldots 1 }}}\equiv 0\pmod{5^{2015}}\\{2015^{2014^{\ldots 1 }}}\equiv 5^{2015}\pmod{4\cdot 5^{2015}}$$
Now lets get back at the second equation
$${2016^{2015^{\ldots 1 }}}\equiv2016^{5^{2015}}\pmod{5^{2016}}$$
By binomial theorem if we write $2016=5\cdot 403+1$ and we notice that $$5^{2015}\mid {5^{2015}\choose k}$$ for $k\not= 0,5^{2015}$ we see that the only term not divisible by $5^{2016}$ is the term $1^{5^{2015}}$ hence we have that
$2016^{5^{2015}}\equiv 1\pmod{5^{2016}}$,now lets combine this equation with the one $\pmod{4}$ and we get ${2016^{2015^{\ldots 1 }}}\equiv 3\cdot 5^{2016}+1\pmod{4\cdot 5^{2016}}$
Now finally we are left with
$$2017^{3\cdot 5^{2016}+1}\equiv 2017\cdot (2017^{5^{2016}})^3\pmod{5^{2017}}$$
Like last time by binomial theorem($2017=403\cdot 5+2$) we can see the only term not divisible by $5^{2017}$ is $2^{5^{2016}}$ so we have that
$$2017\cdot (2017^{5^{2016}})^3\equiv 2017\cdot (2^{5^{2016}})^3\equiv 2017\cdot 8^{5^{2016}}\equiv 2017\cdot 3^{5^{2016}}\pmod{5^{2017}}$$