Proof by induction: Show that $9^n-2^n$ for any $n$ natural number is divisible by $7$. Can someone please solve following problem. 
Show that $9^n-2^n$ for any $n$ natural number is divisible by $7$. ($9 ^ n$ = $9$ to the power of $n$). 
I know the principle of induction but am stuck with setting up a formula for this. 

There shouldn't really be a formula for this problem, just setting up the induction step which usually involves plugging in $n = k+1$ into your expression. First show $7\mid 9^n-2^n$ for $n=1$. Then suppose the result holds for some $k \geq 1$ (i.e. you are allowed to assume $9^k-2^k$ is divisible by $7$.) Consider then the quantity $9^{k+1}-2^{k+1}$. Proving that this quantity must also be divisible by $7$ is all that remains. Doing a little bit of algebra manipulation will get the result. For example notice that  $$9^{k+1}-2^{k+1} = 9^{k}(7+2)-2^{k+1}$$ Can you proceed from here?
