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Show that 9^(n+1)-8n-9 is divisible by 6...

Show that `9^(n+1)-8n-9` is divisible by 64, where `n` is a positive integer.

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In order to showe that `9^(n+1) - 8n -9` is divisible by 64, it has to be proved that `9^(n+1) - 8n - 9 = 64 k`, where k is some natural number
`(1+8)^(n+1) = .^(n+1)C_(0) + .^(n+1)C_(1)(8)+.^(n+1)C_(2)(8)^(2)+"….."+.^(n+1)C_(n+1)(8)^(n+1)`
or `9^(n+1)= 1+(n+1)(8)+8^(2)[.^(n+1)C_(2)+.^(n+1)C_(3) xx 8 + "....."+.^(n+1)C_(n+1)(8)^(n-1)]`
or `9^(n+1)=9+8n+64[.^(n+1)C_(2)+.^(n+1)C+(3) xx 8 + "...."+.^(n+1)C_(n+1)(8)^(n-1)]`
or `9^(n+1) - 8n - 9 = 64 k`, where `k = .^(n+1)C_(2) + .^(n+1)C_(3) xx 8 + "......"+.^(n+1)C_(n+1)(8)^(n-1)` is a positive integers.
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