Is this a legitimate proof? If not, how to prove? Question: Determine all natural numbers $n$ such that: $7 \mid \left(3^n - 2\right)
\implies3^{n}\equiv 2\pmod{7}$
Multiply both sides by 7
$7 \cdot 3^{n}\equiv 7\cdot2\pmod{7}$
Divide both sides by seven, since $\gcd(7,7) = 7$, we have to divide modulus by $7$
$\implies3^{n}\equiv 2\pmod{7/7}$
$\implies3^{n}\equiv 2\pmod{1}$
Therefore $n$ is any natural number, since one divides everything. But I made a mistake somewhere since the original equation doesn't work for $n = 1$

The remainders of $3^n\bmod 7$ starting from $n=1$ are:
$$3,2,6,4,5,1,3,\cdots$$
And by Fermat's little theorem, 
$$3^{n+6}-2 \equiv 3^n-2\pmod 7$$
