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.

$7^{n} +2$ is divisible by $3$ iff $7^{n} +2 - 3 = 7^{n} -1$ is divisible by $3$.
Now, $7^{n} -1 = (7-1)(7^{n-1}+7^{n-2}+\cdots+1)$ is even divisible by $6$.
