<s>[INST] Proving $6^n - 1$ is always divisible by $5$ by induction I'm trying to prove the following, but can't seem to understand it. Can somebody help?

Prove $6^n - 1$ is always divisible by $5$ for $n \geq 1$.

What I've done:
Base Case:
$n = 1$: $6^1 - 1 = 5$, which is divisible by $5$ so TRUE.
Assume true for $n = k$, where $k \geq 1$:
$6^k - 1 = 5P$.
Should be true for $n = k + 1$
$6^{k + 1} - 1 = 5Q$
$= 6 \cdot 6^k - 1$
However, I am unsure on where to go from here.
 [/INST] This is the inductive step written out:
$$
6 \cdot 6^k - 1 = 5 \cdot Q |+1; \cdot \frac{1}{6};-1 \Leftrightarrow 6^k - 1 = \frac{5\cdot Q-5}{6}\underset{P}{\rightarrow}5\cdot P = \frac{5\cdot Q - 5 }{6} | \cdot \frac{1}{5}; \cdot 6\Leftrightarrow Q=6\cdot P + 1
$$
$$
6^k - 1 = \frac{5\cdot Q-5}{6} \overset{Q}{\rightarrow}\ (6^k-1 = \frac{5\cdot (6\cdot P + 1)-5}{6}\Leftrightarrow 6^k-1 = 5\cdot P)
$$
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