Showing $J(A)=0$, where $A$ is the path algebra of an oriented cycle

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Let $Q$ be a cyclic quiver with $n$ vertices and consider the path algebra $A=KQ$. I want to show that the Jacobson radical of $A$ is zero. Since the Jacobson radical is the intersection of all maximal submodules of $A$ I tried to classify them:

For each $1 \leq i \leq n$ we have a maximal submodule of $A$ which is generated by all arrows in $Q$ and all paths of length $0$ except for the path of length $0$ at the $i$-th vertex, so the paths $e_1, \ldots, e _{i-1}, e _{i+1}, \ldots, e_n$. These are maximal submodules in any path algebra. But I was unable to determine any other maximal submodules of $A$. Can someone give me a hint how to find them?

Also is $A$ isomorphic to some other algebraic structure that is easier to work with?