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 $7$.

Simple version without the notation:
$7 \times 1 = 7$
$7 \times 7 = 49$
$7 \times 9 = 63$
$7 \times 3 = 21$
Just look at the last digit in each case.
So the last digit of $7^1$ is $7$.  The last digit of $7^2$ is $9$.  The last digit of $7^3$ is $3$.  And, the last digit of $7^4$ is $1$.
Thus the last digit of $7^5$ is also $7$.  And the last digit of $7^9$ is $7$ (because $7^4 \times 7^4 \times 7 = 7^9$, and the last digits thereof are $1 \times 1 \times 7$.)
And the last digit of $7^{51}$ is the same as the last digit of $7^{47}$, which is the same as the last digit of $7^{43}$, which is the same as the last digit of $7^{39}$ (see the pattern?)...which is the same as the last digit of $7^{7}$, which is the same as the last digit of $7^3$, which is $3$.
By the same logic, the last digit of $7^{1002}$ is the same as the last digit of $7^2$, which is $9$.
