Logarithm as limit Wolfram's website lists this as a limit representation of the natural log:
$$\ln{z} = \lim_{\omega \to \infty} \omega(z^{1/\omega} - 1)$$
Is there a quick proof of this? Thanks

Not a formal proof, only a bit of intuition.
Substituting $x=\ln(z)$ (therefore $z=\exp(x)$), and $n$ instead of $\omega$, this limit is aligned with the infamous Euler's limit for the exponential.
$$ x = \lim_{n \to \infty} n (\exp(x) ^{1/n} - 1) = \lim_{n \to \infty} n (\exp(x/n) - 1 $$
We put the limit aside, and use some trivial algebraic manipulations. Then we put the limit back, to get Euler's limit:
$$\exp(x) = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$$
