Show by contradiction that $a^n\to 0$
Suppose that $\ell > 0$. Then $\frac{\ell}{|a|} > \ell$, hence there is an $N$ such that $|a|^N < \frac{\ell}{|a|}$. But this gives the contradiction $|a|^{N+1} < \ell$.
Suppose that $\ell > 0$. Then $\frac{\ell}{|a|} > \ell$, hence there is an $N$ such that $|a|^N < \frac{\ell}{|a|}$. But this gives the contradiction $|a|^{N+1} < \ell$.