A= (2+1) (2^2+1) (2^4+1) (2^8+1) (2^16+1) (2^32 +1). find the units digit in a- 2016

2^1=2
2^2=4
2^4=16
2^8=256
2^16=65536
2^32=1048576

If we add 1 to the last digits above, we have the sequence
3,5,{7,7,7,7},{7,7,7,7}
But 7*7*7*7=2401, finishes with a 1, so all the groups of 4 7's wont change the last digit.
Hence A ends with a 5 (because 3*5=15 ends with a 5)
and A-2016 ends with a 9  (because 15-6=9)
In fact, A-2016=18446744073709549599  if you bother to know.

Try your deduction skills:
What is the last digit of
B=(2+1) *(2^2+1) *(2^4+1) *(2^8+1) *(2^16+1) *(2^32 +1)...*(2^131072+1)*(2^2097152+1)?
Note: 2097152=2^21

Im pretty sure it would be 6 purple rectangles.