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?