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.
Hint. Let $G$ act on the representation (minus $0$) and use an orbit counting argument to find a copy of the trivial representation.
Let $V$ be a nontrivial irreducible representation. Note that $V$ must be finite dimensional since $G$ is finite. Write $V^\circ = V\setminus\{0\}$ and let $G$ act on $V^\circ$. By orbit stabilizer, $\left| \mathcal{O}_x \right|$ divides $|G|$ for any $x\in V^\circ$, so all $G$-orbits have prime power (more precisely, power of $p$) size. Since these all need to sum to $\left|V^\circ\right|=p^n-1$, there is at least one orbit of size $1$. This orbit is a $G$-invariant one-dimensional subspace, and thus is an isomorphic copy of the trivial representation. But this contradicts that $V$ is irreducible.
Best Answer
It is true (with the caveats given in the other answer): Let $G$ be a finite solvable group, and $p$ a prime dividing the order of $G$. Then the degree of every absolutely irreducible representation (that is, the field sufficiently large) of $G$ in characteristic $p$ divides the order of $G$.
The only proof for this that I am aware of, is a consequence of the Fong-Swan theorem, which states that for solvable groups actually the representation itself must come from reducing a representation in characteristic zero (and then using the result for characteristic zero). This theorem is (not the original reference, but probably the easiest accessible place) Theorem 38 on p. 135 of Serre, Linear Representations of Finite Groups.