**Lemma** $n^{\frac{1}{n}} \to 1$ ad $n \to \infty$. **Proof** If we replace $x$ with $n$ in [[x to the power of the reciprocal of n converges to 1 for any positive x]] we get $h_{n} <1$, which doesn't give us the right limit in the sandwich so we replace $x$ with $\sqrt{n}$ instead. WTS $n^{\frac{1}{2n}} \to 1$. Write $n^{\frac{1}{2n}=}1+h_{n}$ then $\sqrt{n} = (n^{\frac{1}{2n}}){^{n}=(1+h_{n})^{n}} \geq 1+nh_{n} > nh_n$using [[Bernoulli's Inequality]]. Rearranging we have $0<h_{n} \leq \frac{1}{\sqrt{n}}$so by [[Sandwich Rule]], $h_{n} \to 0$ which shows by sum rule that $n^{\frac{1}{2n}} \to 1$. It follows that $n^{\frac{1}{n}}= (n^{\frac{1}{2n}})^{2}\to 1$ by [[Algebra of Limits of Convergent Sequences|product rule for limits]].