How to determine the convergence or divergence of this sequence without resort to the L'Hospital's rule?

Question

Let $$x_n \colon= (n+1)^{ 1 / \ln (n+1) } $$ for all $n \in \mathbb{N}$.

Is the sequence $\left( x_n \right)_{n \in \mathbb{N} }$ convergent in the usual metric space $\mathbb{R}$? If so, how? Can we prove this and also find the limit of this sequence without employing the L'Hospital's rule?

I would like to proceed as rigorously as in an analysis course based on Baby Rudin.

I'm simply clueless on where to begin, what to look for.

Answer 1

Since $n \in \mathbb{N}$, use the exponential identity $a^b = \mathrm{e}^{b \ln a}$. Then the result is evident and immediate.

Answer 2

As suggested by @user296602, consider $\ln x_n$ and use the identity $\ln x^y = y \ln x$.

$$\ln x_n = \ln \left[(n+1)^\frac{1}{\ln(n+1)}\right] = \frac{1}{\ln(n+1)}\ln (n+1) = 1$$

Therefore, since $\exp$ is continuous we conclude $$e = e^{\lim_{n\to\infty} \ln x_n} = \lim_{n\to\infty} e^{\ln x_n} = \lim_{n\to\infty} x_n$$