I get the idea that since $|a|<1$ then for large enough $n$ $a_n<1$ so that when you take the $n^{th}$ power the sequence values become very small. I am struggling to write out the actual $\epsilon-N$ proof however. My inequalities never yield what I need and I keep making mistakes and coming up with inequalities that are not even true.
For example I know that I need $|a_n^n|<\epsilon$ for any given $\epsilon$. Working backwards I could get this if $|a_n|<\epsilon^{1/n}$ but this won't happen because the $a_n^{'s}$ are all close to $a$.
$n\log|a_{n}|\rightarrow-\infty$ since $\log|a_{n}|\rightarrow\log|a|<0$, so $|a_{n}|^{n}=\exp(n\log|a_{n}|)\rightarrow 0$ and hence $a_{n}^{n}\rightarrow 0$.
$\epsilon$-argument:
Given $\epsilon>0$, since $|a|<1$, we can pick an $\eta>0$ so small such that $|a|+\eta <1$. We may take an $N$ so large such that $(|a|+\eta)^{N}<\epsilon$. For the $\eta>0$, we can choose some $n_{0}\geq N$ such that $|a_{n}-a|<\eta$ for all $n\geq n_{0}$, so $|a_{n}|<|a|+\eta$ and hence $|a_{n}|^{n}\leq|a_{n}|^{N}<(|a|+\eta)^{N}<\epsilon$ for all such $n$.