How to find the last one digit of $8^{97}$? What will be the units digit of:
  $8^{97}$


I know I will have to start like:

$8^1$, then $8^2, 8^3$ etc...

is there a shorter method to finding the units/digit?
Hint for proof:

Notice that when you take a product of two numbers the final digit of the result only depends on the final digit of each original factor. Why should that be the case? 

Why does this imply that taking powers of 8 will give a pattern in the final digit? In fact this is true for any natural number, not just 8, and this pattern will always be of a specific sort.
You can use the geometric progression formula which is in the form a*r^(n-1) 

"a" represents the first digit in the sequence. 

"r" represents the ratio between every 2 successive numbers in the sequence

"n" represents the position of the number you are looking for.

So......

a=8^1=1

r=8^2/8^1=8^3/8^2=(well, you get the idea)

r=8

n=97

8*8^(97-1)

Unfortunately I use a 6 bit calculator which will give me the wrong value so just calculate for yourself 8^96 then multiply by8.