Show that $a^1,a^2,a^3,\ldots$ is not bounded from above for $a>1$ I want to prove that the sequence $a^1,a^2,a^3,\ldots$ is not bounded from above for $a>1$?

For any $a>1$ we can always find $n\in\mathbb{N}$ such that $a>1+\frac{1}{n}$ and
$$a^n>\left(1+\frac{1}{n}\right)^n >2$$
$$a^{n} a^{n}>\left(1+\frac{1}{n}\right)^{n} \left(1+\frac{1}{n}\right)^{n} >2\cdot 2$$
$$a^{n} a^{n} a^{n}>\left(1+\frac{1}{n}\right)^{n} \left(1+\frac{1}{n}\right)^{n} \left(1+\frac{1}{n}\right)^{n}>2\cdot 2 \cdot 2$$
$$\ldots$$
So clearly our sequence is not bounded from above.
 A: Your way is reasonable, but uses Bernoulli's inequality. There is another more direct way.
If the sequence is bounded from above, it has a supremum $c$ by the completeness axiom for reals, and by definition we have $a^k > c/a$ for some natural number $k$ because $c/a < c$ since $a > 1$. Thus $a^{k+1} > c$, contradicting the choice of $c$.
Therefore the sequence is not bounded from above.
A: Since $a > 1$, we can write $a = 1 + \delta$ for some $\delta > 0$, by binomial formula, for each $n$, 
$$a^n = (1 + \delta)^n = 1 + n\delta + \cdots + \delta^n > 1 + n\delta.$$
Since $\{n\delta\}$ is unbounded (for fixed $\delta$) by Archimedean property of real numbers, the result follows. 
A: This is another approach which uses proof by contradiction. Suppose that the sequence $s_{n} = a^{n}$ is bounded above. Since $a > 1$ we can see that $$s_{n + 1} = a^{n + 1} = a\cdot a^{n} = as_{n} > s_{n}$$ so that the sequence $s_{n}$ is strictly increasing. Since it is also bounded above it follows that the limit $L = \lim_{n \to \infty}s_{n}$ exists. Moreover because of strictly increasing nature $L$ must be greater than any term in the sequence. In particular $L > a > 1$. Since $s_{n + 1} = as_{n}$ it follows by taking limits that $L = aL$. However this is not possible if both $a , L $ are greater than $1$. This contradiction finishes our job and $s_{n} = a^{n}$ is unbounded.
