Why does $f(z) = z^n$ have no antiderivative only for $n=-1$? The complex valued function $f(z) = z^n$ has an analytic antiderivative on $\mathbb{C} \setminus \{0 \}$ for every $n$ except for $n=-1$. What is so special about $-1$? 
To show why this is such an anomaly, imagine if $z^n$ had an analytic antiderivative on $\mathbb{C} \setminus \{0 \}$ for every $n$ except for $n=3456$. People would demand to know what is so special about $3456$. However, it seems like no one feels the need to explain the anomaly at $n = -1$. What is going on at $-1$? 

When you integrate around the circle,  the little bits $x^k dx$ rotate through $k+1$ multiples of $2\pi$, so they cancel each other out unless $k=-1$.
We want the integral from $x$ to $y$ along one path to be the same as the integral along another path.  Then the integral is just a function of $x$ and $y$.  Imagine putting a path from $y$ to $w$ on the end of that, then the integral from $x$ to $w$ equals the integral from $x$ to $y$ plus the integral from $y$ to $w$.  So the integral is now a function $F(y)-F(x)$.
If two different paths from $x$ to $y$ have the same integral, then the loop from $x$ to $y$ along the first path, and back along the second path, has to be $F(y)-F(x)+F(x)-F(y)=0$.  To repeat, the integral along a closed loop has to be zero.
Suppose a loop can be shrunk to zero, and f(x) is bounded inside the loop.  Break up the area into $N^2$ tiny areas of width $O(1/N)$.  The integral around the large loop equals the sum of the integrals around all the small areas because the inner paths cancel out.  Now we want a condition that makes the integral around a small area to be $O(N^{-3})$.  From memory, that condition is the 'Cauchy Riemann' equations, which is the condition the function f is a function of $z$ and not its conjugate $\overline{z}$.
Now the integral is the sum of $N^2$ integrals, each $O(N^{-3})$, so it is $O(1/N)$.  Let $N$ be very large, and so the integral is zero.
So: The integral around a loop is zero, when $f(z)$ is bounded within the loop, and $f(z)$ does not involve the conjugate.
The only thing left in this case is an integral around zero.  This boils down to a single integral along the unit circle.  So $z=e^{i\theta}$,$dz=e^{i\theta}id\theta$, $z^kdz = ie^{(k+1)\theta}d\theta$, and the result follows.
