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.
For $n\geq 1$, let $S(n)$ denote the statement $$ S(n) : 5\mid(6^n-1)\Longleftrightarrow 6^n-1=5m, m\in\mathbb{Z}. $$ Base case ($n=1$): $S(1)$ says that $5\mid(6^1-1)$, and this is true.
Inductive step: Fix some $k\geq 1$ and assume that $S(k)$ is true where $$ S(k) : 5\mid(6^k-1)\Longleftrightarrow 6^k-1=5\ell, \ell\in\mathbb{Z}. $$ To be proved is that $S(k+1)$ follows where $$ S(k+1) : 5\mid(6^{k+1}-1)\Longleftrightarrow 6^{k+1}-1=5\eta, \eta\in\mathbb{Z}. $$ Beginning with the left-hand side of $S(k+1)$, \begin{align} 6^{k+1} - 1 &= 6^k\cdot 6-1\tag{by definition}\\[0.5em] &= (5\ell+1)\cdot 6-1\tag{by $S(k)$, the ind. hyp.}\\[0.5em] &= 6\cdot 5\ell+5\tag{expand}\\[0.5em] &= 5(6\ell+1)\tag{factor out $5$}\\[0.5em] &= 5\eta.\tag{$\eta=6\ell+1; \eta\in\mathbb{Z}$} \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.
Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
In comments you ask about intuition behind following standard proof of the inductive step $$5\mid 6^k-1\ \Rightarrow\ \bbox[3px,border:1px solid #c00]{\color{#c00}5\cdot 6^k +(6^k\!-1)\,=\, \color{#0a0}{6^{k+1}-1}}\qquad\qquad$$
This is a very natural question since such proofs often appear to be pulled out of a hat, like magic. There is, in fact, a good general explanation for their source. Namely such proofs are simply special cases of the proof of the Congruence Product Rule, as we show below.
${\bf Claim}\rm\qquad\ 6\equiv 1,\, 6^{k}\!\equiv 1 \Rightarrow\ 6^{k+1}\!\equiv 1\ \ \ \pmod{\!5},\ $ a special case of the following
${\bf Lemma}\rm\quad\ \, A\!\equiv a,\, B\!\equiv b\ \Rightarrow\ AB\equiv ab\ \pmod{\!n}\ \ \ $ [Congruence Product Rule]
${\bf Proof}\ \ \ \rm n\mid A\!-\!a,\,\ B-b\,\Rightarrow\, n\mid ( A\!-\!a) B +a\ (B\!-\!b) =A B\,-\,ab$
$\rm\ \ \ \ e.g.\ \ \ 5\mid\ 6\!-\!1,\,\ 6^{k}\!-\!1\ \Rightarrow\ 5\mid\,\bbox[3px,border:1px solid #c00]{(\color{#c00}{6\!-\!1})\,6^{k}+ 1\,(6^{k}\!\!-\!1) = \color{#0a0}{6^{k+1}\!-1}}$
The prior inference is precisely the same as said common proof of the inductive step. Thus we see that this inference is simply a special case of the proof of the Congruence Product Rule. Once we know this rule, there's no need to repeat the entire proof every time we need to use it. Rather, we can simply invoke the rule as a Lemma (in divisibility form if congruences are not yet known). Now the inductive step has vivid arithmetical structure, being the computation of a simple product $\,\rm A,B\equiv 1\Rightarrow AB\equiv 1,\,$ for $\rm\,A\equiv 6,\, B\equiv 6^k.\,$ No longer is the innate arithmetical structure obfuscated by ad hoc divisibility manipulations - since the key idea has been encapsulated into an arithmetically intuitive Lemma (Congruence Product Rule) for convenient reuse.
In much the same way, congruences often allow us to impose intuitive arithmetical structure onto complicated inductive proofs - allowing us to reuse our well-honed grade-school algebra skills manipulating arithmetical equations (vs. more complex divisibility relations). Often introduction of congruence language will serve to drastically simplify the induction, e.g. reducing it to a trivial induction such as $\, 1^n\equiv 1,\,$ or $\,(-1)^{2n}\equiv 1.\,$ The former is the essence of the matter above.
Generally by induction $\,a_i\equiv b_i\,\Rightarrow\, a_1\cdots a_n\equiv b_1\cdots b_n.\,$ The constant case $\,a_i\equiv a,\, b_i\equiv b\,$ yields the ubiquitous Factor Theorem $\,a\equiv b\Rightarrow a^n\equiv b^n$ so $\,a-b\mid a^n=b^n\,$ by choosing the modulus to be $\, n = a-b$.
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 $m$.
This is the inductive step written out: $$ 6 \cdot 6^k - 1 = 5 \cdot Q |+1; \cdot \frac{1}{6};-1 \Leftrightarrow 6^k - 1 = \frac{5\cdot Q-5}{6}\underset{P}{\rightarrow}5\cdot P = \frac{5\cdot Q - 5 }{6} | \cdot \frac{1}{5}; \cdot 6\Leftrightarrow Q=6\cdot P + 1 $$ $$ 6^k - 1 = \frac{5\cdot Q-5}{6} \overset{Q}{\rightarrow}\ (6^k-1 = \frac{5\cdot (6\cdot P + 1)-5}{6}\Leftrightarrow 6^k-1 = 5\cdot P) $$
$6$ 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 $5$.
