<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] We can show by induction that $6^k$ has remainder $1$ after division by $5$.
The base case $k=1$ (or $k=0$) is straightforward, since $6=5\cdot 1+1$.
Now suppose that $6^k$ has remainder $1$ after division by $5$ for $k\ge 1$. Thus $6^k = 5\cdot m+1$ for some $m \in \mathbb{N}$. We can then see that $$6^{k+1}=6\cdot 6^{k} = (5+1)(5\cdot m +1) = 5^2 \cdot m + 5 + 5\cdot m + 1$$
$$=5(5\cdot m + m + 1) + 1.$$
Thus $6^{k+1}$ has remainder $1$ after division by $5$.
Therefore for every $k$, we can write $6^k = 5\cdot m +1$ for some $m$.
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