I have problem proving that these two loops are homotopic
If we have $k_n:I\rightarrow S^1 $ and $k^n :I\rightarrow S^1 $ defined with
$k_{n}(t)=(\cos(2\pi nt),\sin(2\pi nt))$
$k^n(t)=k_1(t) *...*k_1(t)$
$k_1(t)=(\cos(2\pi t),\sin(2\pi t))$
I need to prove that these two loops at $(1,0)$ are homotopic.
Since we are in $S^1$ I can define a homotopy $F:I*I\rightarrow S^1$ with
$F(t,s)=s*k^n(t) +(1-s)*k_n(t)$
I dont know if this is the right map to prove the statement. Thank you very much for helping!
The explicit form of the homotopy would depend on your definition of path concatenation. Since there are infinitely many ways to parametrize the concatenated path I shall give an answer which does not depend on the choice of parametrization.
Suppose you define $\gamma_1\ast\cdots\ast\gamma_n$ by partitioning $I=[0,1]$ into $\{0=t_0\leq t_1\leq\cdots\leq t_n=1\}$ and tracking $\gamma_j$ in the time interval $[t_{j-1}, t_j]$. That is, $(\gamma_1\ast\cdots\ast\gamma_n)(t)=\gamma_j(\frac{t-t_{j-1}}{t_j-t_{j-1}})$ for $t\in[t_{j-1},t_j]$.
$k_n$ is a special case of such formulation; it's when $t_j=\frac{j}{n}$.
Given two different partitions $\{t_0\leq\cdots\leq t_n\}$ and $\{t_0'\leq\cdots\leq t_n'\}$ we can show that they yield homotopic paths. In fact, define a family of intermediate partitions $\{t_0^{(s)}\leq\cdots\leq t_n^{(s)}\}$ by $t_j^{(s)}=(1-s)t_j+st_j'$, where $s\in[0,1]$. Then $F_s(t):=\text{the concatenated path defined by the partition}\;\{t_0^{(s)}\leq\cdots\leq t_n^{(s)}\}$ is the desired homotopy.
The upshot is that through whatever parametrization you define $k^n$, it is homotopic to $k_n$.