Prove $a^n \rightarrow 0$ as $n \rightarrow \infty$ for $\left|a\right| < 1$ without use of log properties $a^n \rightarrow 0$ as $n \rightarrow \infty$ for $\left|a\right|  < 1 $

Hint $u_{2n}$ = $u_{n}^2$
I have totally no idea how to prove this, this looks obvious but i found out proof is really hard... 
I am doing a real analysis course and there's a lot of proving and I stuck there. 
Any advices? Practice makes perfect? 
 A: Replacing $a$ by $|a|$, one can assume without loss of generality that $a$ is a nonnegative real number. If $a=0$, the result is direct. If $0\lt a\lt1$, the sequence defined by $u_n=a^n$ is decreasing and positive hence it converges to some finite nonnegative limit $\ell$. Since $u_{n+1}=au_n$, $\ell=a\ell$. Since $a\ne1$, the only possible limit is $\ell=0$, QED.
The hint that $u_{2n}=u_n^2$ can probably be used as follows, once one knows that the limit $\ell$ exists and is finite: $\ell=\ell^2$ hence $\ell=0$ or $1$ and, since $u_n\leqslant u_1=a\lt1$ for every $n\geqslant1$, $\ell\ne1$ hence $\ell=0$.
A: Proof sketch:  Maybe try showing that $|\frac{1}{a^n}| \to \infty$.
Fix $p = |\frac{1}{a}| > 1$, and let $p = ( 1 + b )$.  Show by induction that $p^n \ge 1 + nb$, and conclude the statement above using the Archimedean property of the reals.
A: Since $0\le|a|\lt1$, we have $0\le|a|^{n+1}\le|a|^n$. Since $|a|^n$ is a non-increasing sequence, bounded below, $A=\lim\limits_{n\to\infty}|a|^n$ exists. Then,
$$
\begin{align}
|a|A
&=|a|\lim_{n\to\infty}|a|^n\\
&=\lim_{n\to\infty}|a|^n\\
&=A
\end{align}
$$
Thus, $(|a|-1)A=0\implies A=0$. Therefore,
$$
\left|\lim_{n\to\infty}a^n\right|=\lim_{n\to\infty}|a|^n=A=0
$$
A: Let $u_n = a^n$ with $|a|< 1$
Let $v_n=|u_n|$
$v_{n+1} = a v_n < v_n$ so $(v_n)$ is decreasing
$0 \le v_n$ so $(v_n)$ is minored
Since $(v_n)$ is minored and decreasing, it converges.

Now let $v_\infty=\lim\limits_{n\to \infty}v_n$
You have $v_{n+1} = a v_n$
By making $n\to \infty$ on both sides (which you can do because $x\mapsto ax$ is continuous) you get:
$v_\infty = a v_\infty$
$0 = (a-1) v_\infty$
But $|a|<1$ so $a-1\not= 0$ so $v_\infty=0$

That is $|u_n|\underset{n\to\infty}{\longrightarrow}0$
So we can conclude that $u_n\underset{n\to\infty}{\longrightarrow}0$
