Consider the following problem:
“Let $I\in \pi_1(S^1,(1,0))$ be the class of the identity map. Show that $nI$ is the class of the map $f_n:S^1\rightarrow S^1$ given by $f_n(z)=z^n$.
I can solve it using the fact that $\pi_1(S^1)=\mathbb Z$. But without assuming this fact, how can one directly solve that problem only using the definition of homotopy? I tried to construct a homotopy writing paths in the polar coordinates but I couldn’t find any.
On
$S^1$ is a topological group under multiplication.
In general if $G$ is a topological group with identity $e$, then there is an alternative multiplication on $\pi_1(G,e)$ given by $[f]\cdot[g]=[h]$ where $h(t)=f(t)g(t)$ (multiplication in the group $G$). The Hilton-Eckmann argument proves that this new multiplication is the same as the standard group operation on $\pi_1(G,e)$ (and is also commutative).
As a consequence, if $f$ is a loop from $e$ to $e$ in $G$, the class of $n[f]$ in $\pi_1(G,e)$ is given by that of the path $t\mapsto f(t)^n$. Your example is $G=S^1$, considered as a subgroup of $\Bbb C^*$.
Let's define loops in $X$ based at $x_0$ to be continuous maps $\gamma\colon[0,1] \to X$ with $\gamma(0) = \gamma(1) = x_0$. This is equivalent to maps $f \colon S^1 \to X$ with $f(1,0) = x_0$, but it's a bit easier for me to think in terms of a real variable.
If $\gamma_1$ and $\gamma_2$ are two loops, the usual definition of their concatenation $\gamma_1 * \gamma_2$ is $$ (\gamma_1 * \gamma_2)(t) = \begin{cases} \gamma_1(2t) & 0 \leq t \leq \frac{1}{2} \\ \gamma_2(2t-1) & \frac{1}{2} \leq t \leq 1 \end{cases} $$ The idea is that you split up the interval $[0,1]$ into two sub-intervals, you do $\gamma_1$ on the first half (scaling $[0,\frac12]$ to $[0,1]$), then $\gamma_2$ on the second half (scaling and shifting $[\tfrac12,1]$ to $[0,1]$).
But you can also break up the interval into unequal pieces $[0,z]$ and $[z,1]$. You would get a weighted-concatenation operation $$ (\gamma_1 *_z \gamma_2)(t) = \begin{cases} \gamma_1\left(\frac{t}{z}\right) & 0 \leq t \leq z \\ \gamma_2\left(\frac{t-z}{1-z}\right) & z \leq t \leq 1 \end{cases} $$ Notice that $\gamma_1 *_{1/2} \gamma_2 = \gamma_1 * \gamma_2$. Also, all of the maps $\gamma_1 *_z \gamma_2$ are homotopic. You just continuously move that breakpoint from one $z$-value to each other. To be explicit: $$ h(t,s) = (\gamma_1 *_{s z_1 + (1-s)z_2} \gamma_2)(t) $$ is a homotopy between $\gamma_1 *_{z_1} \gamma_2$ and $\gamma_1 *_{z_2} \gamma_2$.
Let $\eta_n \colon[0,1] \to S^1$ be the map $\eta_n(t) = e^{2\pi int}$. Notice $\eta_n(t) = f_n(e^{2\pi i t})$, so they induce the same homotopy classes in $\pi_1(S^1)$. Let $m$ and $n$ be positive integers. We want to show that $\eta_m * \eta_n$ is homotopic to $\eta_{m+n}$.
Now $\eta_{m+n}$ completes $m+n$ loops, equally spaced in the interval $[0,1]$. $\eta_m * \eta_n$ also completes $m+n$ loops, but the first $m$ are equally spaced in the interval $[0,\frac12]$, and the last $n$ are equally spaced in the interval $[\frac12,1]$. So let's form $\eta_m *_{m/(m+n)} \eta_n$ instead. It completes its first $m$ loops equally spaced within the interval $[0,\frac{m}{m+n}]$, and its last $n$ loops equally spaced within the interval $[\frac{m}{m+n},1]$. But this means that each loop is completed in the same width interval: $\frac{1}{m}\frac{m}{m+n} = \frac{1}{m+n}$ on the first part, and $\frac{1}{n}\left(1 - \frac{m}{m+n}\right) = \frac{1}{m+n}$ on the second part. So in fact, $\eta_m *_{m/(n+m)} \eta_n = \eta_{n+m}$. Since $\eta_m *_{m/(n+m)} \eta_n$ is homotopic to $\eta_m * \eta_n$, we are done.