Asymptotic complexity of $-\log(c^{1/n} - 1)$
For $c \gt 1$ we have $c^{1/n} - 1 \sim \frac{1}{n}\ln c$, which gives
$$\lim\limits_{n\to\infty}\frac{\log(c^{1/n} - 1)}{-\log(n)} = \lim\limits_{n\to\infty}\frac{\log\ln c - \log(n)}{-\log(n)}=1$$
For proof brought approximation we can consider limit
$$\lim\limits_{x \to 0}\frac{e^x-1}{x} = \lim\limits_{x \to 0}e^x=1$$
based on   L'Hôpital, Taylor expansion etc. De facto it's derivative of exponent in point $0$.