If \(a=1\), then we have a sequence in which every term is \(1\) and the limit equals \(1\).

For \(a>1\) we will show that the limit equals \(\infty\). We use Bernoulli's inequality
\(
(1+x)^n\geq 1+nx.
\)
for \(x=a-1\) and we get
\[
a^n=(1+(a-1))^n\geq 1+n(a-1)\geq n(a-1).
\]
It follows that
\[
\lim_{n\to\infty}n(a-1)=\lim_{n\to\infty}n\lim_{n\to\infty}(a-1)=
\infty(a-1)=\infty.
\]
Using the squeeze lemma for improper limits we get
\(
\lim\limits_{n\to\infty}a^n=\infty.
\)

The remaining case is \(0<a<1\). Evidently \(\tfrac{1}{a}>1\), so we can use the result of the preceding case and we have
\[
\lim_{n\to\infty}a^n=\lim_{n\to\infty}\frac{1}{(\frac{1}{a})^n}=\frac{1}{\lim\limits_{n\to\infty}(\frac{1}{a})^n}=\frac{1}{\infty}=0.
\]