I couldn't quite understand why $S^1$ shouldn't be contractible into one point. I came up with the following homotopy which seems to show that $S^1$ actually is contractible:
Let $x \in S^1$ and $H: I \times S^1 \rightarrow S^1, (r, x) := x^{1-r} \in S^1$. Then $H(x, 0) = x \in S^1$ and $H(x, 1) = 1 \in S^1$.
So, to visualize this: Let $S^1$ be a rope around a pillar for example. If we go with the convention $1 = e^{2\pi i 0}$ and $-1 = e^{2\pi i \frac{1}{2}}$ Standing at one point, the homotopy pulls the rope until it has all the rope.
1) Where is the mistake?
2) Can you tell me (by a mathematical proof, not a visualization) why the sphere is not contractible?
When you take a complex number to a real power, the result may depend on arbitrary choices, and will generally be discontinuous. For example, see if you can describe the function $x\mapsto x^{1/2}$ on $S^1$.
If you want to say it's $e^{i\theta/2}$ for $\theta\in[0,2\pi)$, then as $\theta\to2\pi$ the result approaches $-1$, not $+1$, so the function will not be continuous at $1\in S^1$. Or if you say it's $e^{i\theta/2}$ for $\theta\in(-\pi,\pi]$, then as $\theta\to\pm\pi$ the result approaches the different values $\pm i$, so the function will not be continuous at $-1\in S^1$.
In general, $z^s:=\exp(s\ln z)$, and the logarithm function $\ln$ is where a "branch cut" is chosen.