Limit of $n(a^{1/n}-1)$ as $n \to \infty$ Show that $\lim(n(a^{1/n}-1)) = \ln(a)$

In the context of sequences, I'm not sure how to prove that this is the limit of the sequence. I was trying to convert the expression as follows:
$n(a^{1/n}-1) = \ln(a)\displaystyle\frac{n(e^{\ln(a)/n}-1)}{\ln(a)}$
Then, we can set $\displaystyle\frac{\ln(a)}{n} = 1/x$ to then have:
$\lim(n(a^{1/n}-1)) = \ln(a)\lim({x(e^{x}-1)})=\ln(a)$
Using the result that $\lim(n(e^{n}-1)) = 1$ (both $n$ and $x$ will tend to $\infty$). However, I'm not satisfied with this answer, as I don't think that it is possible to simply set $\displaystyle\frac{\ln(a)}{n} = x$, as then it would be a sequence of in which $x \in \mathbb{R}$, not $x \in \mathbb{N}$ as it should be for a sequence. Is there any other way to solve the problem?
Thank you for your time and patience.

$$\displaystyle \lim_{n \to \infty} n(a^{1/n}-1) = \left . \lim_{x \to 0^+} \frac{(a^x-1)}{x} = \frac{d}{dx}(a^x) \right |_{x=0}=\ln(a)$$
Here, I used the substitution $x=\frac{1}{n}$.
