Although this question has already been asked in general ( How to find $\lim\limits_{n\to\infty} n·(\sqrt[n]{a}-1)$?) , my question is different, because I am stuck with a specific transformation step:
$$\lim\limits_{n\rightarrow \infty} n(\sqrt[n]{a}-1) $$
I've got the solution to it, but I have problems understanding this step: $$\lim\limits_{n\rightarrow \infty} n(\sqrt[n]{a}-1) = \frac{\exp\left(\frac{\log a}{n}\right)-1}{\frac{1}{n}}\\$$
I know that I can rewrite $\sqrt[n]{a}$ as $e^{(1/n)*\log(a)}= \exp\left(\frac{\log a}{n}\right)$ with $\log = \ln$.
Question: How do I get the $\frac{1}{n}$ in the denominator.
And the final step is somehow taking the $\log a$, on both the numerator and denominator.
$$ = \log a \frac{\exp\left(\frac{\log a}{n}\right)-1}{\frac{\log a}{n}}=\log a\\$$
If you are allowed to use the derivative of $a^x$, you can immediately see the definition of $f'(0)$, just set $\frac{1}{n} = t$: $$ \lim_{t \to 0} \lim_{h \to 0} \frac{a^{t+h}-a^h}{t} = \lim_{h \to 0} \lim_{t \to 0} \frac{a^{h+t}-a^h}{t} = \lim_{h \to 0} a^h \log a = \log a $$