Show that $\lim_{n \rightarrow \infty} \sqrt[n]{a} =1$

I think that this one is due to Courant:

For all $n\in\mathbb{N}$ define $a_{n}$ such that $$ a_{n}=\sqrt[n]{n}-1 $$ Note that for every $n\in\mathbb{N}$, $a_{n}\geqslant0$. Now rewrite as: \begin{eqnarray*} n & = & \left(a_{n}+1\right)^{n}\\ & = & \sum_{k=0}^{n}\binom{n}{k}a_{n}^{k}\\ & \geqslant & \binom{n}{2}a_{n}^{2} \end{eqnarray*} We get $n\geqslant\frac{n(n-1)}{2}a_{n}^{2}$.

After rearranging:

$$ a_{n}\leqslant\sqrt{\frac{2}{n-1}}\xrightarrow[n\to\infty]{}0 $$ Concluding that $$ \lim_{n\to\infty}\left(\sqrt[n]{n}-1\right)=\lim_{n\to\infty}a_{n}=0\iff\lim_{n\to\infty}\sqrt[n]{n}=1 $$

Take any number $r>1$. Eventually $r^n>a$ and so eventually $\sqrt[n]a<r$. Now take $r<1$. Eventually $r^n<a$ and so eventually $\sqrt[n]a>r$.

Calculating $\lim\ r^n$ rigorously is a little tricky and depends on how deep you want to go, see this answer for one approach.

I did also use the fact that the $n$-th root function is increasing, but I'm not sure how to prove that without diving into the definition of the ordering in $\mathbb R$.