Find the last Digit of $237^{1002}$? I looked at alot of examples online and alot of videos on how to find the last digit  But the thing with their videos/examples was that the base wasn't a huge number. What I mean by that is you can actually do the calculations in your head. But let's say we are dealing with a $3$ digit base Number... then how would I find the last digit. 
Q: $237^{1002}$
EDIT: UNIVERSITY LEVEL QUESTION. 
It would be more appreciated if you can help answer in different ways. 
Since the Last digit is 7 --> 


*

*$7^1 = 7$

*$7^2 = 49 = 9$

*$7^3 = 343 = 3$ 

*$7^4 = 2401 = 1$ 
$.......$
$........$

*$7^9 = 40353607 = 7$

*$7^{10} = 282475249 = 9$


Notice the Pattern of the last digit. $7,9,3,1,7,9,3,1...$The last digit repeats in pattern that is 4 digits long.


*

*Remainder is 1 --> 7

*Remainder is 2 --> 9

*Remainder is 3 --> 3

*Remainder is 0 --> 1


So, $237/4 = 59$ with the remainder of $1$ which refers to $7$. So the last digit has to be $7$.

$ {\rm mod}\ 10\!:\ \color{#c00}{7^{\large 4}\equiv\bf 1}\,\Rightarrow\, 7^{\large J+4K}\!\equiv 7^{\large J}(\color{#c00}{7^{\large 4}})^{\large K}\!\equiv 7^{\large J}\color{#c00}{\bf 1}^{\large K}\!\equiv 7^{\large J}\, $ by standard Congruence Rules.
Finally write $\ 1002 = J\!+\!4K\ $ for $\,0\le J < 4\ $  and apply the above.
