Why is $\pi_1(\Bbb{R}^n,x_0)$ the trivial group in $\Bbb{R}^n$? My Algebraic Topology book says 

Let $\Bbb{R}^n$ denote Euclidean n-space. Then $\pi_1(\Bbb{R}^n,x_0)$ is the trivial subgroup (the group consisting of the identity alone).

I wonder why that is. I can imagine infinite continuous "loops" in $\Bbb{R}^3$ that start and end at $x_0$.
Thanks in advance!

The problem with your last sentence is that $\pi_1(X,x_0)$ is not the set of loops based on $x_0$, but of homotopy classes of loops based on $x_0$.
Can you see why every loop in $\mathbb R^n$ based on $x_0\in \mathbb R^n$ is homotopic to the constant map based in $x_0$?
