$n$-sphere is simply connected It is supposed that I could proof this proposition with basic knowledge about homotopies and the fundamental group but it doesn't ocurred. Help me please.
I checked similar quetions but proofs are relatively more complicated, and it is supposed according to the book that it must be simpler.

It is of Lee's book Introduction to smooth manifolds, page $614$.
 A: To the best of my knowledge, there are two classic proofs of this fact. One requires you to prove that for any $x\in S^n$ any $f:S^1\to S^n$ is homotopic to a map $f:S^1\to S^n\setminus\{x\}$ (so to speak, you need to "wriggle" your map in a continuous fashion to avoid a set point). This only holds when $n\geq 2,$ but in that case you can use that $S^n\setminus\{x\}$ is homeomorphic to $\mathbb{R}^n$ and since this space is simply connected, the map is null-homotopic.
The second classic proof relies on the Van Kampen Theorem to say that $S^n=(S^n\cap \{(x_j)_{1\leq j\leq n+1}\in \mathbb{R}^{n+1}|x_{n+1}\geq 0\})\cup (S^n\cap \{(x_j)_{1\leq j\leq n+1}\in \mathbb{R}^{n+1}|x_{n+1}\leq 0\})$ and both of these are simply connected (they are, again, homeomorphic to $\mathbb{R}^{n}$) and their intersection is path connected when $n\geq 2$.
A: Here is a sketch of an elementary proof. We will use the following facts: 
$1).$ It suffices to prove that if $f : [0, 1] \to S^n$ is a loop in $S^n$, it is null-homotopic.
$2).\ S^n$ with a point removed, is homoeomorphic to $\mathbb R^n$.
$3).$ So, $S^n$ with a two points removed, is homoeomorphic to $\mathbb R^n$ with a point removed.
Let $U=S^n\setminus \{(0,\ \cdots, 1)\}$ and $V=S^n\setminus \{(0,\ \cdots, -1)\}.$ It is no loss of generality to suppose that $f$ is a loop whose starting (and ending) point is not $(0,\cdots, 1$),because if it is, then we can repeat the argument below, with $(0,\cdots,-1)$ in place of $(0,\cdots,1).$ 
Now, $f^{-1}(U)$ and $f^{-1}(V)$ form a cover of $[0,1]$, so the Lebesgue number lemma applies to show that there is a partition $0<x_1<\cdots <x_{n-2}<1$ such that $f$ maps $[x_i,x_{i+1}]$ into either $U$ or $V$ (or both, but if this is the case, choose one of $U$ or $V$.) Letting $f_i$ denote the restriction of $f$ to $[x_i,x_{i+1}]$, divide $I=\{1,2,\cdots, n\}$ into two sets of indices: the first, $I'$, consists of those $i's$ for which $f_i$ maps into $U$, and the second, $I''$, consists of those $i's$ for which $f_i$ maps into $V$. 
If $i\in I'$, then $f_i$ does not pass through $(0,\cdots, 1)$ but if $i\in I''$, consider $V\setminus \{(0,\cdots , 1)\}.$ By $3).,$ this set is homeomorphic to $\mathbb R^n$ with a point removed, which is path-conncected. Thus, there are continuous functions $f'_i:[x_i,x_{i+1}]\to V\setminus \{(0,\cdots , 1)\}$ such that $f_i(x_i)=f'_i(x_i)$ and $f_i(x_{i+1})=f'_i(x_{i+1}).$ Now $2).$ implies that $f_i\sim f'_i$ via some homotopy $H_i:[x_i,x_{i+1}]\times I\to V.$ Composing with the inclusion map, and still calling the new homotopy $H_i$ we get $H_i:[x_i,x_{i+1}]\times I\to S^n.$
It follows that $g=f_i\cup f'_i:[0,1]\to S^n$ is a continuous loop that misses $(0,\cdots, 1)$ and satisfies $g(0)=f(0)$ and $g(1)=f(1).$ Then,
$H(s,t)=\begin{cases}
f_i(s)\quad s\in [x_i,x_{i+1}];\ i\in I'\\
H_i(s,t)\quad s\in [x_i,x_{i+1}];\ i\in I''
\end{cases}$
is a homotopy from $f$ to $g$ and $g$ misses $(0,\cdots, 1)$; that is, $g$ maps into $U$. Now $2).$ implies that $g$ is null-homotopic, and so, so is $f$. 
