Prove that these two loops are homotopic 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!
 A: 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 $k_n$.
