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. 

Assume $7\mid(9^n-2^n)$; then you can write $9^n-2^n=7m$, for some integer $m$, or $9^n=7m+2^n$ as well. Then
$$
9^{n+1}-2^{n+1}=
9\cdot 9^n-2^{n+1}=
9(7m+2^n)-2^{n+1}=
63m+(9-2)2^n=
7(9m+2^n)
$$
