number theory - How to prove that $4^n-3n-1$ is divisible by 9?

How can I prove that $4^n-3n-1$ is divisible by $9$? I tried dividing the expression by $9$ and seeing if the terms cancelled in any predictable way but I still cannot prove it. Maybe there is a clever solution but so far I have been unable to spot it.

Answer

Call $a_n=4^n-3n-1$. Then $$a_{n+1}=4^{n+1}-3n-4=4(4^n-3n-1)+9n$$ so two consecutive terms differ by a multiple of $9$. Since $a_0=0$ is divisible by $9$, all of them are.