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 $7$.

We show that the last two digits of $237^{1002}$ are $69$.
For any $n \in \mathbb N$, the two digits of $n$ are given by $n\bmod{100}$. If $\gcd(n,100)=1$, $n^{\phi(100)}=n^{40} \equiv 1\pmod{100}$, by Euler's theorem.
Since $\gcd(237,100)=1$, applying Euler's theorem gives
$$ 237^{1002} = \big(237^{40}\big)^{25} \cdot 237^2 \equiv 237^2 \equiv 37^2 \equiv 69\pmod{100}. $$
This proves our claim. $\blacksquare$
