If $f$ is a closed path in $S^1$ at $1$ and if $m \in \Bbb Z$, then $t \rightarrow f(t)^m$ is a closed path in $S^1$ at $1$ and $\deg f^m = m \deg f$
It's easy to see that they are both closed paths in $S^1$ at $1$ since they're essentially elements of $\{ z \in \Bbb C : |z| = 1\}$, but I'm having trouble showing the second part.
I can show by induction and by constructing a homotopy that $\deg f^m = m \deg f$, for $m \in \Bbb Z^+$ through use of the fact that $d : \pi_1(S^1, 1) \rightarrow \Bbb Z$ is an isomorphism with $[f] \rightarrow \deg f = f'(1)$, where $f'(1)$ is the lifting of $f$ into $(\Bbb R, t_0)$ where $t_0$ is an integer.
But I'm confused as to how I'd show this for $m=0$ or $m \in \Bbb Z^-$.
Is there any easier way to prove this, or does anyone have any idea how to show their equality with negative numbers?
Let $\theta\colon\mathbb{R}\rightarrow\mathbb{R}$ be a lift of $f$ through the universal cover $p\colon t\mapsto e^{2i\pi t}$, then notice that $m\theta$ is a lift of $f^m$. Indeed, notice that: $$f^m(t)=\left(e^{2i\pi\theta(t)}\right)^m=e^{2i\pi m\theta(t)}.$$ Therefore, $\deg(f^m)=m\theta(1)-m\theta(0)=m(\theta(1)-\theta(0))=m\deg(f)$.