Prove if $a>1$ then $\lim_{n\rightarrow\infty}a^{n}=\infty $ Good morning i was thinking about this problem and I make this. I need someone review my exercise and say me if that good or bad. Thank!

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

Proof:
Suppose $\left\{ a^{n}\right\} $ is 
monotonically increasing. In other words $a^{n}<a^{n+1}<
 a^{n+2}...$ and  Suppose $\left\{ a^{n}\right\} $ is Bounded set then $\left\{ a^{n}\right\} $ converge. By definition   $\lim_{n\rightarrow\infty}a^{n}=L$. 
We know this
$\left(a^{n+1}-a^{n}\right)=a^{n}(a-1)$ , $(a-1)>0$. Because $a>1$
Then
$a^{n}(a-1)>(a-1)\Rightarrow a^{n}>1$
Exist $N ∈ \mathbb{N} $ such that $a^{N}$ > $L$ and $\left\{ a^{n}\right\} $ is non bounded set Then $\left\{ a^{n}\right\} $ diverge and 
$\lim_{n\rightarrow\infty}a^{n}=\infty$

But, i don't sure it is fine, please help.

An other way
Let $a>1$. $$a^n=e^{n\ln(a)}\underset{\ln(a)>0}{>}n\ln(a)\underset{n\to \infty }{\longrightarrow }\infty .$$
An other way (using Bernoulli)
Since $a>1$, there is $\varepsilon>0$ s.t. $$a=1+\varepsilon.$$
Using Bernoulli,
$$a^n=(1+\varepsilon)^n\geq n\varepsilon+1.$$
