I am looking for the simplest proof that $\mathbb{R}^n$ is simply connected, i.e. its fundamental group is trivial (only one element). Everywhere I looked, this thing is thought too easy to prove, so that a direct proof is skipped. People only focus on n-spheres and other topologically interesting objects.
Thank you!
Suppose $\gamma : [0, 1] \to \mathbb{R}^n$ is a loop based at $x_0$. Let
$$\Gamma(s, t) = t \cdot x_0 + (1-t) \cdot \gamma(s)$$
Then $\Gamma : [0, 1] \times [0, 1] \to \mathbb{R}^n$ is continuous, $\Gamma(s, 0) \equiv \gamma(s)$ and $\Gamma(s, 1) \equiv x_0$. Also for each $t_0 \in [0, 1]$ the curve $\Gamma(s, t_0)$ is a loop based at $x_0$. Hence $\Gamma$ is a homotopy of $\gamma$ and a constant loop.
That means that every loop is homotopic with a constant loop, which means that the fundamental group is trivial.
Additional remark (thanks to qbert in the comments): the same proof can be applied to show that every convex subset $S \subseteq \mathbb{R}^n$ is also simply connected.