Proof of $\lim_{n\to \infty} \sqrt[n]{n}=1$

Here is one using $AM \ge GM$ to $1$ appearing $n-2$ times and $\sqrt{n}$ appearing twice.

$$\frac{1 + 1 + \dots + 1 + \sqrt{n} + \sqrt{n}}{n} \ge n^{1/n}$$

i.e

$$\frac{n - 2 + 2 \sqrt{n}}{n} \ge n^{1/n}$$

i.e.

$$ 1 - \frac{2}{n} + \frac{2}{\sqrt{n}} \ge n^{1/n} \ge 1$$

That the limit is $1$ follows.

Since $x \mapsto \log x$ is a continuous function, and since continuous functions respect limits: $$ \lim_{n \to \infty} f(g(n)) = f\left( \lim_{n \to \infty} g(n) \right), $$ for continuous functions $f$, (given that $\displaystyle\lim_{n \to \infty} g(n)$ exists), your proof is entirely correct. Specifically, $$ \log \left( \lim_{n \to \infty} \sqrt[n]{n} \right) = \lim_{n \to \infty} \frac{\log n}{n}, $$

and hence

$$ \lim_{n \to \infty} \sqrt[n]{n} = \exp \left[\log \left( \lim_{n \to \infty} \sqrt[n]{n} \right) \right] = \exp\left(\lim_{n \to \infty} \frac{\log n}{n} \right) = \exp(0) = 1. $$

Here's a two-line, completely elementary proof that uses only Bernoulli's inequality:

$$(1+n^{-1/2})^n \ge 1+n^{1/2} > n^{1/2}$$ so, raising to the $2/n$ power, $$ n^{1/n} < (1+n^{-1/2})^2 = 1 + 2 n^{-1/2} + 1/n < 1 + 3 n^{-1/2}.$$

I discovered this independently, and then found a very similar proof in Courant and Robbins' "What is Mathematics".