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.

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.
