We have to prove that $9^{n+1}$ - 8n - 9 = 64k ∴ $9^{n+1}$ - 8n - 9 = $(8+1)^{n+1}$ - 8n - 9 [put 9 = 8 + 1] = [$\binom{n+1}{8}^{n+1} + \dots + \binom{n+1}{n-2} 8^{3}$ + $\binom{n+1}{n-1} 8^{2} + \binom{n+1}{n} 8 + \binom{n+1}{n+1}$] - 8n - 9 = $\binom{n+1}{8}^{n+1}$ + ... + $\binom{n+1}{n-2} 8^{3}$ + $\binom{n+1}{n-1} 8^{2}$ + (n + 1) 8 + 1 - 8n - 9 = $\binom{n+1}{8}^{n+1}$ + ... + $\binom{n+1}{n-2} 8^{3}$ + $\binom{n+1}{n-1} 8^{2}$ = $8^{2}$ [$\binom{n+1}{0} 8^{n-1} + \dots + \binom{n+1}{n-2} 8$ + $\binom{n+1}{n-1}$] = 64k [Where , k = $\binom{n+1}{0} 8^{n-1}$ + ... + $\binom{n+1}{n-1}$] Hence, $9^{n+1}$ - 8n - 9 is divisible by 64, whenever n is a positive integer.