Using the binomial expansion , we get

`(1+x)^(n+1) = .^(n+1)C_(0)+.^(n+1)C_(1)x+.^(n+1)C_(2)x^(2)+.^(n+1)C_(3)x^(3)+...+.^(n+1)C_(n+1)x^(n+1).`

Putting x=8 in the above expansion , we get

`9^(n+1) = 1 +(n+1) xx 8+.^(n+1)C_(2) xx (8)^(2) + .^(n+1)C_(3) xx (8)^(3)+...+.^(n+1) C_(n+1) xx(8)^(n+1)`

`rArr (9^(n+1)-8n-9)=(8)^(2) xx{.^(n+1)C_(2)+.^(n+1)C_(3) xx8+...+(8)^(n-1)}

`rArr (9^(n+1)-8n-9)=64 xx " (an integer) "`.

Hence, `(9^(n+1)-8n-9) " is exactly divisible by " 64.`