Prove that for every positive integer n, 9^n-8n-1 is divisible by 64.

This question screams proof by induction, so we start with the base case, which in this case is n=1:

9^1-8-1 which is indeed divisible by 64.

Now, let’s assume that it holds true for some positive integer n=k. ie:

9^k-8k-1=64p for p\in\mathbb{Z}.

Now let’s see how we can use this to prove that the statement holds true for n=k+1. For n=k+1 we have:

9^(k+1)-8(k+1)-1=9(9^k)-8k-8-1=9(9^k-8k-1)+64k

where we have manipulated the expression to contain the left hand side of the inductive hypothesis. Thereby, plugging in the inductive hypothesis, we get:

9^(k+1)-8(k+1)-1=9(9^k)-8k-8-1=9(64p)+64k=64(9p+k)

but clearly 9p+k is an integer, so this is divisible by 64 and thus the statement holds true for n=k+1, thus it holds true for all positive integers k

How clear is this post?