Show that there always exist a $n \in \mathbb {N}$ with $a,b \in \mathbb {R}$ and $b>1$ such that $b^n>a$.

Question

I am not allowed to use limits. I have to prove it by exploiting the Bernoulli inequality.

If $b\geq a$, then there is nothing to show. If $b <a$ and if $b\geq2$ then I can say $$a <[a]+1 <1+(b-1)([a]+1)<b^{[a]+1}$$ I don't know how to argue in the case $2>b>1$

Answer 1

But by Bernoulli $$b^n=(1+b-1)^n\geq1+n(b-1)>a$$ for $$n>\frac{a-1}{b-1}.$$

Answer 2

Set $b=1+x$ , $x>0$, real.

$b^n = (1+x)^n \ge 1+nx$, $n \in \mathbb{N}.$

Archimedean principle:

For $r$, real, there is a $n$ , $n \in \mathbb{N}$, s.t.

$n>r.$

Choose $r=\dfrac{a-1}{x}$.

There is a $n> \dfrac{a-1}{x},$ or

$1+nx >a.$

Finally:

$b^n = (1+x)^n \ge 1+nx >a.$