The formula
$$\oint \frac{dz}{z^{n+1}}=2\pi i\delta_{n,0}$$
A special case of Cauchy's integral formula, clearly seems to be related to the factoid that $\int dz/z = \ln(z)$, considering the "exception" for the $1/z$ case.
But is there an intuitive explanation of this relation?
On
Set $z=e^{i\theta}$ and you'll get
$$\oint\frac{\mathrm dz}{z^{n+1}}=\int_0^{2\pi}\frac{\mathrm d(e^{i\theta})}{e^{(n+1)i\theta}}=i\int_0^{2\pi}e^{ni\theta}~\mathrm d\theta=\frac{e^{2\pi ni\theta}-1}n$$
For any $n\in\mathbb Z\setminus0$, this equals zero, and for $n\to0$, this equals $2\pi i$ (definition of the derivative of $e^x$)
If $n\neq0$, then $\frac1{z^{n+1}}$ has a primitive and therefore its integral along a closed path is equal to $0$.
But I suppose that it is the case $n=0$ that interests you. Well, define $\log(z)=\log(r)+i\theta$, whenever $z=r\bigl(\cos(\theta)+i\sin(\theta)\bigr)$ ($\theta\in(-\pi,\pi)$). Then $\log'(z)=\frac1z$ and therefore$$\oint\frac1z\,\mathrm dz=i\pi-(-i\pi)=2\pi i.$$Of course, this is not a proof. It just provides a connection between this integral and the $\log$ function.