Trying to show that $\ln(x) = \lim_{n\to\infty} n(x^{1/n} -1)$ How do I show that $\ln(x) = \lim_{n\to\infty} n (x^{1/n} - 1)$?
I ran into this identity on this stackoverflow question. I haven't been able to find any proof online and my efforts to get from $\ln(x) := \int_1^x \frac{\mathrm dt}t$ to that limit have been a failure.

Starting with the indicated integral, first note that for any $0<a<b$
$$
\int_{a}^b\frac{dt}t=\int_{a}^b\frac{d(at)}{at}=\int_1^{b/a}\frac{dt}t
$$
from which the logarithm identity $\ln(b)-\ln(a)=\ln(b/a)$ follows.
Per induction $\ln(a^n)=n\ln(a)$ and thus 
$$
\ln(x)=n\ln(\sqrt[n]x)=n\int_1^{\sqrt[n]x}\frac{dt}t=n\left(\sqrt[n]x-1\right)\frac1{1+\theta_n(\sqrt[n]x-1)},\quad \theta_n\in(0,1)
$$
the last by the mean value theorem in integral form. Thus
$$
n\left(\sqrt[n]x-1\right)=\ln(x)·\left(1+\theta_n(\sqrt[n]x-1)\right)
$$
and as $\sqrt[n]x\to 1$ for $n\to\infty$, the right side converges to $\ln(x)$.

Note that these calculations avoid any knowledge about properties and derivatives of the exponential function.
