Let $n\in\Bbb N_{0}$ and consider the immersion $$\phi:\Bbb S^1\to \Bbb R^2:e^{it}\mapsto (\cos(nt),\sin(nt))$$ Prove that the degree of the Gauss map of $\phi$ is equal to $n$. How is this compatible with the proof of the Gauss-Bonnet Theorem using the Gauss map.
Preliminaries:
We always consider $\Bbb R^m$ endowed with the euclidean metric, denoted by $g_{0}$ (the corresponding norm is denoted by $\Vert\cdot\Vert$).
If $\varphi:(M,g)\to\Bbb R^{n+1}$ is an isometric immersion from an $n$-dimensional oriented manifold $(M,g)$ into $\Bbb R^{n+1}$, the Gauss map of $\varphi$ is the application $$\nu: M\to \Bbb S^n$$ such that if $x\in M$ and $\{e_{1},\dots,e_{n}\}$ is an oriented basis of $T_{x} M$, then $\{e_{1},\dots,e_{n},\nu(x)\}$ is an oriented basis of $\Bbb R^{n+1}$ ; and such that $\nu(x)\perp T_{\varphi(x)}\varphi(M)$ along with $\Vert \nu(x)\Vert=1$.
Remark: this definition makes no sense to me. I assume that they mean "if $x\in M$ and $\{e_{1},\dots,e_{n}\}$ is a basis of $T_{\varphi(x)}\varphi(M)$, then $\{e_{1},\dots,e_{n},\nu(x)\}$ is a basis of $T_{\varphi(x)}\varphi(M)$". I understand that maybe identifications are made but this is very confusing for me.
The degree $\text{deg}(f)$ of an application $f:M\to N$ is defined as follows. Let $n\in N$ be such that any element of $f^{-1}(\{n\})=:R(f,n)$ is a regular value of $f$, i.e. a point of $M$ where the differential of $f$ does not vanish. Then, for each $m\in R(f,n)$, the index $\text{ind}_{m}(f)$ of $f$ at $m$ is defined as $1$ if $f_{\ast_{m}}$ is orentiation-preserving and $-1$ else. Then,
$$\text{deg}(f) := \sum_{m\in R(f,n)}\text{ind}_{m}(f)$$
Firstly, as I am not given any metric, I suppose I need to use the induced metric by the euclidean metric $g_{0}$ defined on $\Bbb R^2$, i.e.
$$g_{e^{it}}=\phi^{\ast}g_{0}\vert_{\phi(e^{it})}$$
I suppose I can take the atlas on $\Bbb S^1$ given by
$$\mathcal{A}=\{(U_{1},\psi_{1}),(U_{2},\psi_{2})\}$$
where
\begin{align*} &U_{1}=\{e^{it}\mid -\pi<t<\pi\}\\ &U_{2}=\{e^{it}\mid 0<t<2\pi\} \end{align*}
and $\psi_{j}:U_{j}\to\Bbb R:e^{it}\mapsto t$ ($j=1,2$). A local coordinate on $\Bbb S^1$ is given by $x^{j}=\psi_{j}$ ($j=1,2$). Therefore, the induced metric is given by
$$g_{e^{it}}=n^{2}dt\wedge dt$$
Indeed, if $x:N\to \Bbb R^{m}:(u_{1},\dots,u_{n})\mapsto (x^{1}(u),\dots,x^{m}(u))$ is an immersion ($n<m$), then
$$x^{\ast}g_{0}=\sum_{j=1}^{m}(\sum_{i=1}^{n}\frac{\partial x^{j}}{\partial u_{i}}du_{i})^{2}$$
Computation of the Gauss map
Now, I need to compute the Gauss map $\nu$ of $\phi$. Here, we can take
$$\nu(e^{it})=(\cos(nt),\sin(nt))$$
which is indeed an element of $\Bbb S^1$ (where $\Bbb S^1$ seen as a subspace of $\Bbb R^2$). To check it is indeed orthogonal to $T_{\phi(e^{it})}\Bbb \phi(S^{1})$, I take the curve $$(-1,1)\to\Bbb S^{1}:s\mapsto\gamma(s)=(\cos(n(s+t)),\sin(n(s+t)))$$ which obviously goes through $\phi(e^{it})$ at $s=0$ and compute its derivative at $s=0$, which is $\dot\gamma(0)=(-n\sin(nt),n\cos(nt))$. This vector spans the tangent space and is orthogonal to $\nu(e^{it})$. Moreover, $\{\dot\gamma(0),\nu(e^{it})\}$ obviously spans $\Bbb R^{2}$ as they are orthogonal non-zero vectors.
I am not sure how to check the orientation.
From these observations, I suppose that if the manifold $M,g$ is a sphere with an isometric immersion, then the immersion is the Gauss map.
Why cannot I get the right degree?
Let $t\in(0,2\pi)$. Then, for a point $(\cos(nt),\sin(nt))$ on the sphere,
$$\nu^{-1}\{(\cos(nt),\sin(nt))\}=\{e^{it}\}$$
But then, obviously, the degree cannot be $n$, so I must have made a mistake but cannot see where.
Moreover, I don't get how the question about Gauss-Bonnet is relevant since it only applies to even dimensional manifolds and $\Bbb S^1$ is obviously of dimension $1$.
EDIT: After a comment from Michał Miśkiewicz, it is now clear that I made a mistake calculating the preimage of $(\cos(nt),\sin(nt))$. The right one is
$$\{e^{it+\frac{2\pi ik}{n}}\mid k=1,\dots,n\}$$