Proof by induction that $5|11^n-6$ for all positive integers $n$ 
Prove by induction that $5|11^n-6$ for all positive integers $n$

Let $p(n) = 11^n-6.$ We have $p(1) = 5$, thus it holds for $p(1)$. Assume it holds for $p(k)$. We will prove that it's true for $p(k+1)$. We have $p(k+1) = 11^{k+1}-6$. But $6 = 11^k-f(k).$ Thus
$$f(k+1) = 11\cdot 11^k-11^k+f(k) = 10\cdot 11^k+f(k). $$
$5|(10\cdot 11^k+f(k))$ by hypothesis thus the statement is true for $p(k+1)$. Hence it's true for $p(n)$. 
Is the above correct?

Suppose $5\mid 11^n-6\implies 11^n-6=5k$ for some $k\in\mathbb{Z}$.
\begin{align*}
11(11^n-6)&=11\cdot5k \\
11^{n+1}-11\cdot 6+60&=11\cdot 5k+60 \\
11^{n+1}-6&=11\cdot 5k+60 \\
11^{n+1}-6&=5(11k+12)
\end{align*}
$$11^{n+1}-6=5(11k+12)\implies 5\mid 11^{n+1}-6$$
So we have shown
$$5\mid 11^n-6\implies 5\mid 11^{n+1}-6$$
Check $5\mid 11^1-6$, and conclude $\forall n\in\mathbb{N},5\mid 11^n-6$.
