Monotone sequence proof 
If $a$ satisfies $0<a<1$, show that the sequence $X=(a^n)$ is
  convergent by proving that it is monotone, decreasing and bounded.

My attempt:
Since $a$ satisfies $0<a<1$ then $\implies$ $0^n<a^n<1^n= 0<a^n<1$ which then implies that $1<\frac{1}{a^n}$ so $0<a^n<1<\frac{1}{a^n}$ which then implies $0<a^{2n}<a^n<1$. How can I finish this up or is this wrong?
 A: Hint: If $a<1$ then $xa< ?$ for any positive $x$.
Apply this inductively to $a^k$ to show the sequence is monotone decreasing.
A: It is clear that the sequence $(a^n)$ is bounded below by $0$. But observing it is bounded below by, say, $-17$ would be enough for the proof of convergence.
To show the sequence is monotone decreasing, we need to show that $a^{n+1}\lt a^n$ for any non-negative integer $n$.  This is easy, for $a^{n+1}=a^n\cdot a\lt 1\cdot a^n =a^n$.
And we know that any monotone decreasing sequence that is bounded below has a limit.
Remark: In the post, a full proof was not given for the fact that the sequence is monotone decreasing. It was only shown that $a^{2n}\lt a_n$.
We can now prove that the limit of our sequence is $0$. For let $L$ be the limit. Then 
$$\lim_{n\to\infty}a^{n+1}=\lim_{n\to\infty} a^n=L.$$
But also 
$$\lim_{n\to\infty} a^{n+1}=\lim_{n\to\infty}a\cdot a^n=a\lim_{n\to\infty}a^n=aL.$$
It follows that $L=aL$ and therefore $L=0$.
This proof used some "limit laws." We can also show that the limit is $0$ by a direct "$\epsilon$-$N$" argument.
