Find the last digit of $22^{23^{24^{25^{26^{27^{28^{\cdot^{\cdot^{\cdot^{67}}}}}}}}}}$ Find the last digit of $22^{23^{24^{25^{26^{27^{28^{\cdot^{\cdot^{\cdot^{67}}}}}}}}}}$ in base-$10.$
Just to clarify, I want to find the last digit of the power tower of consecutive numbers starting at $22$ and ending at $67.$ I was just wondering if there was some general method that could be used to solve this?
I don't know if Euler's Theorem might be useful for solving this problem, but it seems there might be a very simple approach.
 A: Note that $\ \color{#c00}{23^{\large 24^{\Large 25^{\ \cdot^{\ \cdot^{\ \cdot}}}}}\!\!\!\! =\, 1\!+\!4N}\ $ by $\bmod 4\!:\ 23^{2K}\!\equiv (-1)^{2K}\equiv 1$
$\begin{align} \Rightarrow\ 22^{\smash[t]{\color{#c00}{23^{\large 24^{\Large 25^{\ \cdot^{\ \cdot^{\ \cdot}}}}}}}}\!\!\!\!\!\!\!\!\!\bmod 10   &\,=\, 22^{\color{#c00}{1+4N}}\bmod 10,\ \ \text{so factoring out $2$ by }\rm\color{#90f}{MDL}\ (below) \\[.4em]  
&\,=\, 2\! \underbrace{(\color{#0a0}2^{\color{#0a0}4N} 11^{1+4N})}_{\textstyle\equiv  \color{#0a0}1^N\   1^{\,1+4N}\!\!\pmod{\!5}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!}\!\! \bmod 5\, =\, 2\end{align}$ 
using:  $\ ab\bmod ac\,=\, a(b\bmod c),\,$ the $\!\bmod\!$ Distributive Law $ = \rm\color{#90f}{MDL}$
Remark $ $ We chose to evaluate the $\rm\color{#c00}{exponent} \bmod 4\,$ because after pulling out the factor $\,2\,$ via MDL we are left with modulus $5,\,$ and $\, a^4\equiv 1\pmod {\!5}\,$ when $\,5\nmid a\,$ (by Fermat or Euler or directly). Therefore it suffices to consider the exponent $\bmod 4\,$ upon applying modular order reduction. Applied recursively, the same idea suffices to handle arbitrary modular exponent towers.
A: Considering the exponent of $23$, we see that it is a multiple of $4$, and that $23^{4n}\equiv 1\mod{8}$. So the whole expression is $22^k$ where $k\equiv 1\mod{8}$, and thus the last digit is $2$.
