The question of how to classify all finite-dimensional representations of $C_n$ over an arbitrary field $F$ can be studied using the structure theorem for finitely-generated modules over a principal ideal domain, in this case $F[x]$. The structure theorem asserts that any finitely-generated module is uniquely a finite direct sum of modules of the form $F[x]/p(x)^r$ where $p \in F[x]$ is irreducible and $r$ is a non-negative integer.
If $T$ is an operator acting on $F^k$ for some $n$, then $F^k$ becomes a finitely-generated module over $F[x]$ with $x$ acting by $T$. $T$ gives a representation of the cyclic group $C_n$ if and only if $T^n = 1$, in which case the summands $F[x]/q(x)^r$ in the decomposition of $F^k$ must have the property that $q(x)^r | x^n - 1$.
If $F$ has characteristic $0$ or has characteristic $p$ and $p \nmid n$, then $x^n - 1$ is separable over $F$, hence $r \le 1$ and $F^k$ is a direct sum of irreducible representations, all of which are of the form $F[T]/q(T)$ where $q$ is an irreducible factor of $x^n - 1$ over $F$.
If $F$ has characteristic $p$ and $p | n$, then writing $n = p^s m$ where $p \nmid m$ we have
$$x^n - 1 = (x^m - 1)^{p^s}$$
from which it follows that $r \le p^s$ (but now it is possible to have $r > 1$). If $r > 1$, then the corresponding representation $F[T]/q(T)^r$ is indecomposable and not irreducible, where $q$ is an irreducible factor of $x^m - 1$ over $F$. The irreducible representations occur precisely when $r = 1$. In other words,
The irreducible representations of $C_{p^s m}$, where $p \nmid m$, over a field of characteristic $p$ all factor through the quotient $C_{p^s m} \to C_m$.
One can also see this more directly as follows. If $V$ is an irreducible representation of $C_{p^s m}$ over a field of characteristic $p$ and $T : V \to V$ is the action of a generator, then
$$T^{p^s m} - 1 = (T^m - 1)^{p^s} = 0.$$
Thus $T^m - 1$ is an intertwining operator which is not invertible, so by Schur's lemma it is equal to zero.
Consider a regular representation of $G$ i.e. $G$ acts on the left on the group $\mathbb{F}$-algebra $F[\mathbb{G}]$. This is a faithful finite dimensional representation. Therefore, the class of finite dimensional faithful representations of $G$ is nonempty. Pick now a faithful and finite dimensional representation $\rho:G\rightarrow \mathrm{GL}(V)$. Decompose $V$ as a direct sum of irreducible representations
$$V = V_1\oplus V_2\oplus ...\oplus V_k$$
of $G$ (this requires $|G|$ coprime to characteristic of $\mathbb{F}$). For every $i=1,2,...,k$ consider $N_i\subseteq G$ the set of $g\in G$ such that $\rho(g)$ acts trivially on $V_i$. Then $N_i$ is a normal subgroup of $G$ (it is a kernel of the induced representation $\rho_{\mid V_i}$). Moreover, since $\rho$ is faithful, we derive that
$$\bigcap_{i=1}^kN_i = \{1\}$$
Let $N$ be a minimal (with respect to inclusion) nontrivial normal subgroup of $G$. If each $N_i$ is nontrivial, then $N\subseteq N_i$ and hence
$$N\subseteq \bigcap_{i=1}^kN_i$$
This is a contradiction and hence there exists $i_0\in \{1,2,...,k\}$ such that $N_{i_0}= \{1\}$. Therefore, $\rho_{\mid V_{i_0}}$ is irreducible and faithful.
Best Answer
Hint. Let $G$ act on the representation (minus $0$) and use an orbit counting argument to find a copy of the trivial representation.