real analysis - Proof that $lim_{nrightarrow infty} sqrt[n]{n}=1$

Thomson et al. provide a proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$ in this book. It has to do with using an inequality that relies on the binomial theorem. I tried to do an alternate proof now that I know (from elsewhere) the following:

$$\begin{align} \lim_{n\rightarrow \infty} \frac{ \log n}{n} = 0 \end{align}$$

Then using this, I can instead prove:

$$\begin{align} \lim_{n\rightarrow \infty} \sqrt[n]{n} &= \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \newline & = \exp{0} \newline & = 1 \end{align}$$

On the one hand, it seems like a valid proof to me. On the other hand, I know I should be careful with infinite sequences. The step I'm most unsure of is:

$$\begin{align} \lim_{n\rightarrow \infty} \sqrt[n]{n} = \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \end{align}$$

I know such an identity would hold for bounded $n$ but I'm not sure I can use this identity when $n\rightarrow \infty$.

If I am correct, then would there be any cases where I would be wrong? Specifically, given any sequence $x_n$, can I always assume:

$$\begin{align} \lim_{n\rightarrow \infty} x_n = \lim_{n\rightarrow \infty} \exp(\log x_n) \end{align}$$

Or are there sequences that invalidate that identity?

(Edited to expand the last question)
given any sequence $x_n$, can I always assume:

$$\begin{align} \lim_{n\rightarrow \infty} x_n &= \exp(\log \lim_{n\rightarrow \infty} x_n) \newline &= \exp(\lim_{n\rightarrow \infty} \log x_n) \newline &= \lim_{n\rightarrow \infty} \exp( \log x_n) \end{align}$$
Or are there sequences that invalidate any of the above identities?

(Edited to repurpose this question).
Please also feel free to add different proofs of $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$.

Answer

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

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