For $|r| < 1$ the sequence $r^k$ is null, and for $|r| > 1$ the sequence $|r|^k$ diverges to $+\infty$

First, $|r^k| = |r|^k$ for every $k \in \mathbb{N}$, by induction: at $k = 0$ both sides are $|1| = 1$, and if $|r^k| = |r|^k$ then $|r^{k+1}| = |r^k r| = |r^k|\,|r| = |r|^k |r| = |r|^{k+1}$.

Case zero. Assume $r = 0$.

Case small. Assume $0 < |r| < 1$.

Case large. Assume $|r| > 1$.

In case zero, $r^k = 0$ for every $k \ge 1$: indeed $r^1 = r^0 r = 1 \cdot 0 = 0$, and if $r^k = 0$ then $r^{k+1} = r^k r = 0$, so induction gives the claim from $k = 1$ on.

In case small, put $s := 1/|r|$, which is defined since $|r| \ne 0$, and $h := s - 1$. Then $s > 1$ and $h > 0$.

In case large, put $h' := |r| - 1$, so $h' > 0$ and $|r| = 1 + h'$.

In case zero, for every rational $\varepsilon > 0$ and every $k \ge 1$ we have $|r^k - 0| = |0| = 0 < \varepsilon$, so $r^k \to 0$ and claim 1 holds.

In case small, $|r|^k s^k = (|r| s)^k = 1^k = 1$, so $|r|^k = 1/s^k$, and $s^k > 0$.

In case small, Bernoulli applied to $h > 0 \ge -1$ gives $s^k = (1+h)^k \ge 1 + kh > kh > 0$ for every $k \ge 1$, using $1 > 0$ and $kh > 0$.

In case large, Bernoulli applied to $h' > 0 \ge -1$ gives $|r|^k = (1 + h')^k \ge 1 + k h'$ for every $k \in \mathbb{N}$.

In case large, let $M \in \mathbb{R}$ be arbitrary and use [L7] to fix a natural $n \ge 1$ with $M/h' < n$; then $M \le n h'$, since multiplying $M/h' \le n$ by $h' > 0$ preserves the inequality.

In case small, let $\varepsilon > 0$ be rational; then $\varepsilon h > 0$, so [L7] supplies a natural $N \ge 1$ with $1/N < \varepsilon h$, whence $1/(Nh) \le \varepsilon$ on multiplying by $1/h > 0$.

In case small, combining steps 3.2 and 3.3: $0 < kh < s^k$ gives $|r|^k = 1/s^k < 1/(kh)$ for every $k \ge 1$.

In case large, for every $k \ge n$ we have $k h' \ge n h' \ge M$, so $|r|^k \ge 1 + k h' \ge 1 + M > M$, the last step because $1 > 0$.

In case small, for every $k \ge N$ we have $kh \ge Nh > 0$, hence $1/(kh) \le 1/(Nh) \le \varepsilon$, and therefore $|r^k - 0| = |r^k| = |r|^k < 1/(kh) \le \varepsilon$.

In case large, an index $n$ has been produced for an arbitrary real $M$ with $|r|^k > M$ for all $k \ge n$, which is exactly divergence to $+\infty$: claim 2 holds.

In case small, the rational $\varepsilon > 0$ was arbitrary and the index $N$ was produced from it, so $r^k \to 0$ and claim 1 holds.

The hypothesis $|r| < 1$ of claim 1 is exhausted by cases zero and small, since $|r| \ge 0$ with $|r| = 0$ exactly when $r = 0$, so trichotomy leaves only $0 < |r| < 1$; the hypothesis $|r| > 1$ of claim 2 is case large. Both claims are therefore established.