Prove if $0Good morning i have a proof of this exercise but i don't know if my proof is correct. please review that!

Problem:
Prove if $0<a<1$ then $\lim_{n\rightarrow\infty}a^{n}=0  $

Proof:
Let $0<a<1$ then 
$$a^{n}=e^{\ln(a)^{n}}=e^{n\ln(a)}>n\ln(a)$$
$$\lim_{n\rightarrow\infty}n\ln(a)=\ln(a)\lim_{n\rightarrow\infty}n=-\infty$$
And then:
$$\lim_{n\rightarrow\infty}e^{n\ln(a)}=e^{-\infty}=0.$$

Now, it's correct my proof? Thanks.
 A: Fix $a \in (0,1)$ and define $b=1/a>1$. Then
$$
a^n=e^{n \ln a}=e^{n \ln b^{-1}}=e^{-n \ln b}.
$$
Now
$$
b>1 \implies \lim_{n\to \infty}{n \ln b} = \infty.
$$
Therefore, since $\mathrm{exp}$ is a continuous function, we conclude
$$
\lim_{n\to \infty} a^n=\lim_{n\to \infty}e^{-n \ln b}=e^{-\lim_{n\to \infty}n \ln b}=0.
$$
A: The sequence $a^n$ is positive, decreasing and bounded below by $0.$ Therefore it converges to some $L \in [0,1).$ It follows that $a\cdot a^n \to a\cdot L.$ But $a\cdot a^n = a^{n+1},$ so it also converges to $L.$ Therefore $aL = L.$ If $L>0,$ we get $a=1,$ contradiction. Therefore $a^n\to 0.$ 
A: Another way:
$\displaystyle \sum_{k=1}^{\infty} a^k$ is a geometric series that converges for $a \in (0, 1)$.  The convergence of this series implies $\displaystyle \lim_{n \rightarrow \infty} a^n = 0$.
A: let $\varepsilon >0,0<\left| a \right| <1$ using Bernoulli inequation we have $$\\ \frac { 1 }{ { \left| a \right|  }^{ n } } ={ \left( 1+\left( \frac { 1 }{ { \left| a \right|  }^{ n } } -1 \right)  \right)  }^{ n }>1+n\left( \frac { 1 }{ { \left| a \right|  } } -1 \right) >n\left( \frac { 1 }{ { \left| a \right|  } } -1 \right) \\ { \left| a \right|  }^{ n }=\left| { a }^{ n } \right| <\frac { \left| a \right|  }{ n\left( 1-\left| a \right|  \right)  } <\varepsilon ,\forall n>\frac { \left| a \right|  }{ \varepsilon \left( 1-\left| a \right|  \right)  } $$
