show that \left| \frac{a}{b} \right| <1 then \lim_{k\rightarrow \infty} \left( \frac{a}{b} \right)^k=0

I'm questioning my proof because of the absolute value but here's what I have:

0< \left| \frac{a}{b} \right| <1 let q= \frac{a}{b} then 0< |q| <1

let \delta >0 and |q|=\left| \frac{1}{1+\delta} \right|

now applying the Bernoulli inequality I get:

0 <|q^k| = \left| \frac{1}{(1+\delta)^k} \right|\leq  \left| \frac{1}{1+k\delta} \right|

\forall \ k \ \in \mathbb{N}, \ \epsilon >0 \ \mbox{let} \ k_0 \ \in \mathbb{N}: \frac{1}{k_0} <k\epsilon \Rightarrow k\geq k_0

0 <|q^k| = \left| \frac{1}{k\delta} \right| \leq  \left| \frac{1}{k_0\delta} \right| \leq \left| \frac{1}{\delta} \delta\epsilon \right| =|\epsilon|

therefore \left(\frac{a}{b} \right)^k \rightarrow 0 as n\rightarrow \infty