<s>[INST] Prove that $7^n+2$ is divisible by $3$ for all $n ∈ \mathbb{N}$ 
Use mathematical induction to prove that $7^{n} +2$
  is divisible by $3$ for all $n ∈ \mathbb{N}$.

I've tried to do it as follow. 
If $n = 1$ then  $9/3 = 3$.
Assume it is true when $n = p$. Therefore $7^{p} +2= 3k $ where $k ∈ \mathbb{N} $. Consider now $n=p+1$. Then 
\begin{align}
&7^{p+1} +2=\\
&7^p\cdot7+ 2=\\
\end{align}
I reached a dead end from here. If someone could help me in the direction of the next step it would be really helpful. Thanks in advance.
 [/INST] Show that $7^n\equiv 1 (\text{mod} 3)$
$(6 + 1)(6 + 1) = 36 + 12 + 1$
$(6 + 1)(6 + 1)(6 + 1) = 216 + 72 + 6 + 36 + 12 + 1$
So all $(6 + 1)^n$ involve the sum of a series of terms which are multiples of $6$ (divisible by $3$) except for the last term which is $1$
Hence $7^n\equiv 1 (\text{mod} 3)$
and $7^n + 2$ is always divisible by $3$
</s>