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.
 A: Hint: $(7^{p+1}+2)-(7^p+2)=6\times7^p$.
A: Hint
If $7^n+2=3k$ then $$7^{n+1}+2=7(\color{red}{7^n})+2=7(\color{red}{3k-2})+2.$$
A: Step: n+1.
$7^{n+1}+2= (6+1)7^n+2=$
$ 6\cdot 7^n +(7^n+2).$
The first term is divisible by $3$, as is the second term (hypothesis).
A: 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 $3$
A: Conceptually the inductive step gets $\,P(n\!+\!1)\,$ by scaling $\,P(n)\,$ by $\,\color{#c00}{7 \equiv\,1}\pmod{\!3}\,$ using CPR = Congruence Product Rule. If congruences are unknown we can preserve this arithmetical essence by using an analogous product rule for divisibility as below, where  $\ m\mid n\ $ means $\,m\,$ divides $\,n.\,$
$$\qquad\qquad\qquad\!\! \begin {align}
&3\mid\  \color{#c00}{7\ \ - \,\  1}\\
&3\mid\ \ \  7^{\large n} -\ B\qquad\ P(n)\qquad\ [B = -2\,\ {\rm in\  OP}]\\
\Longrightarrow\ \ &3\mid\ \color{#c00}7 7^{n}\! -\color{#c00}1B\qquad P(n\!+\!1)
\end{align} $$ 
$\begin{align}{\bf Divisibility\ Product\ Rule}\ \ \ \  
&m\mid\  a\ -\ b\qquad {\rm i.e.}\quad \   a\,\equiv\, b\\
&m\mid \ \ A\: -\: B\qquad\qquad \ A\,\equiv\, B\\
\Longrightarrow\ \ &\color{}{m\mid aA - bB}\quad  \Rightarrow\quad aA\equiv bB\!\pmod{\!m}\\[.2em]
{\bf Proof}\,\ \  m\mid (\color{#0a0}{a\!-\!b})A + b(\color{#0a0}{A\!-\!B}) &\,=\, aA-bB\ \ \text{by $\,m\,$ divides $\rm\color{#0a0}{green}$ terms by hypothesis.}\end{align}$
The point of using congruences (vs. divisibilities) is that it allows us to reuse our strong intuition about operations (vs. relations), e.g. the above congruence product rule is the analog of multiplying the equations $a =b,\ A = b\,$ to get $\,aA = bB,\,$ something which is less intuitive arithmetically when expressed in divisibility language. This will be clarified when one studies ring theory - where one learns the relationship between congruences and quotient rings (here $\Bbb Z\bmod n\,\cong\, \Bbb Z/n\Bbb Z$) 
You can find further discussion of these topic many prior posts.
A: $7^{n} +2$ 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 $6$.
