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.

$6$ has a nice property that when raised to any positive integer power, the result will have $6$ as its last digit. Therefore, that number minus $1$ is going to have $5$ as its last digit and thus be divisible by $5$. 
