I'm trying to see if this is true, but not sure how to go about it. I know that $f$ induces a map on chains where $C_1(S^1) \to C_1(S^1)$ is given by $\sigma \mapsto f\circ \sigma$.
How would I go about showing that $f_*: H_1(S^1) \to H_1(S^1)$ is given by multiplication by $n$?
On
I'd use the argument principle to show that the map $f(z) = z^n$ encircles the origin $n$ times as $z = \mathrm{e}^{\mathrm{i}\theta}$ orbits the origin once (i.e. as $\theta$ runs through $[0,2\pi]$). (If needed, you might also show that $|f(z)| = 1$.)
Now we know the generator of the abelian group $\langle c \rangle = H_1(S^1)$ is sent to $f_*(c) = n c$. As a consequence, we know that, for all $k \in \Bbb{Z}$, $f_*(kc)= nkc$. That is, $f_*$ is the multiply by $n$ map. (You can also show this directly from the argument principle: $\mathrm{e}^{\mathrm{i}k\theta}$ encloses the origin $k$ times while $f(\mathrm{e}^{\mathrm{i}k\theta})$ encloses the origin $nk$-times).
I'll give a method which uses only the Eilenberg-Steenrod axioms and makes no specific reference to singular cohomology. This means that the same statement is true for your favourite generalised cohomology theory (K-theory, cobordism, $BP$-theory, etc...). It also has the upside that it is topological and quite visual, rather than relying on algebraic manipulations.
To begin, we'll notice that $S^1\subseteq\mathbb{C}$ has a multiplication $\mu:S^1\times S^1\rightarrow S^1$ which turns it into a topological group. The degree $2$ map is then the composite $$\underline 2:S^1\xrightarrow{\Delta}S^1\times S^1\xrightarrow{\mu}S^1$$ where $\Delta$ is the diagonal map $z\mapsto (z,z)$. Check directly that $\underline 2$ is exactly the map $z\mapsto z^2$. To get the degree $n+1>2$ map we have inductively $$\underline {(n+1)}:S^1\xrightarrow{\Delta}S^1\times S^1\xrightarrow{\underline n\times 1}S^1\times S^1\xrightarrow{\mu}S^1.$$ For negative $n$ we use the group inverse $z\mapsto z^{-1}$. The identity is the degree $1$ map, and the constant map is the degree $0$ map.
I'll focus from here on the case for $n=2$. The method will make it clear that an inductive step (which I'll leave to you) gives the answer for all $n\in\mathbb{Z}$.
To proceed recall that the wedge, or one-point union, $S^1\vee S^1$ is the quotient space of the disjoint union $S^1\sqcup S^1$ which is obtained by identifying together a single point in each summand. It may be described geometrically as a figure $8$. There is an obvious map $$c:S^1\rightarrow S^1\vee S^1$$ obtained by pinching together the points $\pm1$. Note that in homology it induces a map $$c_*:H_1(S^1)\rightarrow H_1(S^1\vee S^1)\cong H_1(S^1)\oplus H_1(S^1).$$
Now, in the category of abelian groups, finite coproducts and finite products coincide (i.e. a finite direct sum of abelian groups is isomorphic to the finite direct product of the same groups). The point is that the map $c_*$ is determined completely by its composites with the two maps $$H_1(S^1)\xleftarrow{pr_1}H_1(S^1)\oplus H_1(S^1)\xrightarrow{pr_2}H_1(S^1)$$ which project onto each of the factors.
But with respect to the isomorphism $H_1(S^1\vee S^1)\cong H_1(S^1)\oplus H_1(S^1)$, these projections are just the homomorphisms induced by the maps $$S^1\xleftarrow{q_1}S^1\vee S^1\xrightarrow{q_2} S^1$$ which crush one side of the figure $8$ to a point and act as the identity on the other. You can check quite easily that there are homotopies $$q_1\circ c\simeq id_{S^1}\simeq q_2\circ c,$$ and, using the homotopy invariance of homology, conclude that $c_*$ is the homomorphism $$c_*:x\mapsto x\oplus x.$$
Now, using the same homotopies you just wrote down, you can check that the composite $$S^1\xrightarrow{c}S^1\vee S^1\hookrightarrow S^1\times S^1$$ is homotopic to the diagonal map $$S^1\rightarrow S^1\times S^1,\qquad z\mapsto (z,z).$$ It is also clear that the composite $$S^1\vee S^1\hookrightarrow S^1\times S^1\xrightarrow{\mu}S^1$$ is the fold map $\nabla:S^1\vee S^1\rightarrow S^1$, which is the identity on each factor.
The point is that the degree $2$ map I wrote down at the beginning is homotopic to the composite $$S^1\xrightarrow{c}S^1\vee S^1\xrightarrow{\nabla}S^1.$$ Thus to compute $\underline 2_*:H_1S^1\rightarrow H_1S^1$ it suffices to compute the action of the composite $$H_1(S^1)\xrightarrow{c_*}H_1(S^1)\oplus H_1(S^1)\xrightarrow{\nabla_*}H_1(S^1).$$ We computed $c_*$ above. To get $\nabla_*$ we observe that not only does the wedge axiom tell you that there is an isomorphism $$H_1(S^1)\oplus H_1(S^1)\xrightarrow{\cong}H_1(S^1\oplus S^1)$$ but it also tells you how this isomorphism is induced. Namely, it says that this isomorphism is induced by the sum of the maps induced by the inclusions $$S^1\xrightarrow{in_1} S^1\vee S^1\xleftarrow{in_2}S^1.$$ Using this observation, we check directly that $\nabla_*$ is the map $$\nabla_*:H_1(S^1)\oplus H_1(S^1)\rightarrow H_1(S^1),\qquad (x,y)\mapsto x+y.$$
Putting everything together, we have that $\underline{2}_*: H_1(S^1)\rightarrow H_1(S^1)$ is the map $(\nabla\circ c)_*=\nabla_*\circ c_*$, which is exactly the map $$x\mapsto x\oplus x\mapsto x+x=2x.$$ i.e., it is multiplication by $2$. Cue induction.