The units digit of a power tower of consecutive numbers, from 2019 to 1 
Is it possible to find the units digit of
  $2019^{2018^{2017^{.^{.^{.^{3^{2^{1}}}}}}}}$?

Where the expression contains all natural numbers $[1,2018]$ as powers and $2019$ as the main base.
Any help would be appreciated. THANKS!
 A: HINT
$2019\equiv -1\pmod{10}\Rightarrow 2019^a\equiv (-1)^a\pmod{10}$
A: The unit digit of a number $N$ can be computed as the class of $N$ modulo $10$ and taking its representant in $\{0,1,...,9\}$
You're asking for the units digit of a number of the form $2019^M$. Since the class modulo $10$ of $2019$ is $-\overline1$, you are in fact computing $(-1)^M\bmod 10$.
The result is $1$ if $M$ is even and $9$ if $M$ is odd.
Since your actual $M$ is of the form $M=2018^m$ and $2018$ is even, the former case holds.
A: $2019^{2018^{2017^{.^{.^{.^{3^{2^{1}}}}}}}}\equiv9^{2018^{2017^{.^{.^{.^{3^{2^{1}}}}}}}}\pmod{10}$
As $\phi(10)=4,(9,10)=1$ and the exponent is multiple of $4$
$9^{2018^{2017^{.^{.^{.^{3^{2^{1}}}}}}}}\equiv9^0\pmod{10}$
A: Hint:
Have a look at the unit digits of $9,9^2,9^3,9^4,9^5,\dots$.
Would things be different if we switch $9$ with $2019$?
A: HINT : Assume $a_n=n^{n-1^{n-2^{.^{.^{.^{3^{2^{1}}}}}}}} \mod 10$, then the sequence will be :
$$
1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0...
$$
A: If we wanted to be tedious we could note that $$2019\equiv -1\bmod 10\\2018\equiv 2\bmod 4\\2017\equiv 1\bmod 2\\2016\equiv 0\bmod 1$$
and then cut the power tower down to size, by first noting the exponents above the 0 don't have an effect, and that exponent 0 leads to 1 which is useless as an exponent, so we have $-1^2\equiv 1$  as our last digit.  EDIT if you wanted to have to figure out the whole tower, you'd need to ask for the last 674 digits. 
