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.
 A: 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,
$$a^n=(1+\varepsilon)^n\geq n\varepsilon+1.$$
A: Simply use this version of Bernoulli's inequality:

For any $a>0$, one has $\quad a^n-1\ge n(a-1)$

to show than $a^n$ can be made larger than any prescribed number.
A: You can achieve your proof using contradiction: If the limit $L$ of $(a^n)$ is finite then 
$$\lim_{n\to\infty} a^{n+1}-a^n=0=\lim_{n\to\infty} a^n(a-1)=L(a-1)\implies  L=0$$
which is a contradiction.
An alternative proof is: let $h>0$ such that $a=1+h$ so
$$a^n=(1+h)^n\ge 1+nh\xrightarrow{n\to\infty}+\infty$$
A: What you need to show is that for any $x > 0$, there is an $N > 0$ such that $a^N > x$.
Given that you have shown the difference between elements is greater than $a - 1$, this means that you can choose $N$ to be any integer larger than $\frac{x}{a-1}$.
A: A problem I see with your proof is that 
$$
(a-1)>0
$$
does NOT implies that
$$
a^n(a-1)>(a-1).
$$
This means that you cannot just cancel $(a-1)$ on both sides to get $a^n>1$. Anyway, the fact can be easily proved by induction so this is not the real problem here.
The main problem with your proof is that you seemed to claim that if $a^n>1$ for all $n\in\Bbb N$, then there exists an $N\in\Bbb N$ such that 
$$
a^N>L\ .
$$
This does not follow logically from your previous points. 
If you somehow think that I misunderstood you, you'll have to be more explicit in each of your steps.
A: Your phrasing is very bad — for instance, you say 'suppose $\{a^n\}$ is monotonically increasing', but in fact that's not something you should suppose; it's something that's true, and you should prove it.
You've got a good idea in the next step: once you know that $\{a_n\}$ is monotonic increasing, you can use the Monotone Convergence theorem to derive a contradiction.  Unfortunately, as others have noted your attempted proof from there is mathematical gibberish.
Instead, you can use an argument like this: Suppose that $\{a^n\}$ were bounded.  Then by the Monotone Convergence theorem it has a limit $L$.  By the definition of limit, for every $\epsilon$ we can choose an $n_0$ such that $|L-a^n|\lt\epsilon$ for all $n\gt n_0$.  Now, the plan is to find a particular $\epsilon$ for which this breaks down.  If we pick some specific $m\gt n_0$ and the 'right' epsilon, then the idea is that $a^m$ is 'close enough' to $L$ that $a^{m+1}=a\cdot a^m$ is guaranteed to be larger than $L$ (what size does $\epsilon$ have to be for you to guarantee this?); then $a^{m+2}=a\cdot a^{m+1}\gt aL$.  But this implies that $|a^{m+2}-L|\gt (aL-L)=(a-1)L$, and if $\epsilon$ is chosen correctly, then this supplies the contradiction and proves that the  assumption of boundedness must be wrong.
