[Math] Prove that $2^n3^{2n}-1$ is always divisible by 17

Question

Prove that $2^n3^{2n} -1$ is always divisible by $17$.

I am very new to proofs and i was considering using proof by induction but I am not sure how to. I know you have to start by verifying the statement is true for the integer 1 but I dont know where to go from there.

Answer 1

$$2^n\cdot3^{2n}-1 = 18^n-1 = (18-1)(\cdots) = 17(\cdots)$$