Proof verification.Limit of sequence by definition.

Question

Prove that $lim_{n\to\infty} {q^n} =0$ if $|q|<1$. I need prove that: $$(\forall \epsilon >0 )(\exists n_0=n_0(\epsilon))(\forall n \in \mathbb{N})(n\geq n_0 \implies |q^n|<\epsilon)$$ Let $\epsilon >0$ arbitrarily. $$|q^n|=|q|^n< \epsilon$$ Now use logarithm with base $|q|$ and choose $n_0> log_{|q|}\epsilon$. I think that this is correct but in books they use some inequality and i understand their proof also but i only need verification of my solution.