I am trying to prove that for any $\displaystyle b \in \mathbb{R}$ such that $\displaystyle b > 1$. We have that for any $\displaystyle n \in \mathbb{Z}^+$, $\displaystyle b^{\frac{1}{n}}$ can be made arbitrarily close to 1.

I'm completing the routine outlined in Rudin's book. So I'm at the step where I need to prove: $\displaystyle b - 1 \geq n(b^{\frac{1}{n}} - 1)$. And I have this so far:

$\displaystyle b=(b^{\frac{1}{n}})^{n}=((b^{\frac{1}{n}}-1)+1)^{n}$. Which I expand by the binomial theorem to get:

$\displaystyle \sum_{i=0}^{n}{n \choose k}(b^{\frac{1}{n}}-1)^{n-k}1^{k}={n \choose n}(b^{\frac{1}{n}}-1)^{n}+{n \choose 1}(b^{\frac{1}{n}}-1)^{n-1}+\cdots+{n \choose n-1}(b^{\frac{1}{n}}-1)+1.$

Where my plan was to drop all but the last two terms by claiming that they are non-negative and therefore:$\displaystyle b - 1 \geq n(b^{\frac{1}{n}} - 1)$. However to claim they are non-negative I need to use the fact that $\displaystyle b^{\frac{1}{n}} > 1$. Which is the fact I'm trying to prove.

So can anyone see what I'm doing incorrectly at this point? Thanks in advance.