answer to heureka's comment=

8 also divides 64

you may argue that 4, 2, 1 also divides 64

but i have already proved it above

answer to melody's comment=

when we take 8 common from (nc_{0}8^n +nc_{1}8^n-1 + ...........+nc_{n}8^0) we get (nc_{0}^8n-1,nc_{1}8^n-2....)

for eg- 8 common from (1(nc_{0}8^n) we get= 8(8^n-1)

nc_{1}8^n-1=8(nc_{1}8^n-2) etc

remaining same

what do you think

please reply back

exams are approaching

and binomial theorem carries 90 marks