<s>[INST] Proving $9^n - 8n - 1$ is divisible by $8^2$ for $n\ge 0$? My textbook provided the following proof:


  
*Base case: When $n=0, 9^n-8n-1=0=64\cdot0$, so $64\mid\left(9^n-8n-1\right)$.
Induction step: Suppose that $n\in\mathbb N$ and $64\mid\left(9^n-8n-1\right)$. Then there is some integer $k$ such that $9^n-8n-1=0=64k$. Therefore:
$$\begin{align}
9^{n+1}-8(n+1)-1&=9^{n+1}-8n-9\\
&=9^{n+1}-72n-9+64n\\
&=9\left(9^n-8n-1\right)+64n\\
&=9(64k)+64n\\
&=64(9k+n)
\end{align}$$
so $64\mid\left(9^{n+1}-8(n+1)-1\right)$.

What I don't understand is how the equation goes from 
$$= 9^{n+1} - 8n - 9$$
to 
$$= 9^{n+1} - 72n - 9 + 64n$$
 [/INST] Express $8n$ as $72n-64n$. We hence have
$$9^{n+1} - {\color{red}{8n}} - 9 = 9^{n+1} - \color{red}{(72n-64n)} - 9 = 9^{n+1} \color{red}{- 72n + 64n} - 9 = 9^{n+1} - \color{red}{72n} - 9 + \color{red}{64n}$$
</s>