 Solution

Use the allowed Ringel lemma on bricks in the following precise form: a finite-dimensional indecomposable quiver representation that is not a brick module contains a nonzero brick $B$ with ${\mathrm{Ext}}_Q^1(B,B)\ne 0$.

Suppose such a nonbrick indecomposable existed, and let $\mathbf{b}=\dim B$. The Tits form of a quiver is $q_Q(\mathbf{b})=⟨\mathbf{b},\mathbf{b}{⟩}_Q$, so the Ringel form identity would givePositive definiteness gives $q_Q(\mathbf{b})>0$ for nonzero $\mathbf{b}$, a contradiction. Therefore every indecomposable is a brick.