= Solution
For a finite-dimensional algebra over an <algebraically closed field>, <basic algebra> means $A/J(A)\cong k^n$, so its <simple modules> are one-dimensional and occur once in the semisimple quotient of the regular <module>. A <hereditary ring> has every <submodule> of a <projective module> projective; equivalently its <global dimension> is at most one.
Choose representatives $S_1,\ldots,S_n$ of the <simple modules>. Our <Ext quiver> convention gives one vertex $i$ for $S_i$ and
$$
\#\{i\to j\}=\dim_k\operatorname{Ext}^1_A(S_i,S_j).
$$
With primitive orthogonal <idempotents> $e_i$, put $P_i=Ae_i$ and $J=J(A)$. Applying $\operatorname{Hom}_A(-,S_j)$ to $0\to JP_i\to P_i\to S_i\to0$ gives
$$
\operatorname{Ext}^1_A(S_i,S_j)
\cong\operatorname{Hom}_A(JP_i/J^2P_i,S_j).
$$
The restriction map from $\operatorname{Hom}(P_i,S_j)$ is zero since every map to a simple kills $JP_i$, and $P_i$ is projective. Thus the arrow multiplicity is $\dim e_j(J/J^2)e_i$. This specifies the orientation rather than silently replacing the <quiver> by its opposite.
Because $A$ is hereditary, $JP_i$ is projective. Its semisimple top has $a_{ij}$ copies of $S_j$, where $a_{ij}$ is the arrow multiplicity. Its <projective cover> therefore supplies an epimorphism
$$
\bigoplus_jP_j^{a_{ij}}\twoheadrightarrow JP_i.
$$
This splits by projectivity of $JP_i$. Its <kernel> is a summand contained in the radical of the source; projection onto that summand and <Nakayama lemma> force the <kernel> to be zero. Hence
$$
\boxed{JP_i\cong\bigoplus_jP_j^{a_{ij}},\qquad
\dim P_i=1+\sum_ja_{ij}\dim P_j.}
$$
Every arrow $i\to j$ forces $\dim P_i>\dim P_j$, so $Q$ has no directed cycles.
Choose lifts in $e_jJe_i$ of bases of the arrow spaces. Send the vertices and arrows of the <path algebra> to these <idempotents> and lifts, multiplying paths in the order $\beta\alpha$ for consecutive arrows $\alpha:i\to j$, $\beta:j\to l$. This defines an <algebra homomorphism> $kQ\to A$. It is surjective: vertices span $A/J$, arrows span $J/J^2$, and products show that paths of length $r$ span $J^r/J^{r+1}$. Nilpotence of $J$ then gives all of $A$.
Let $N_i$ be the number of directed paths starting at $i$, including the length-zero path. Acyclicity gives the same recursion $N_i=1+\sum_ja_{ij}N_j$, with value one at a sink. Induction from sinks gives $N_i=\dim P_i$. Therefore
$$
\dim kQ=\sum_iN_i=\sum_i\dim P_i=\dim A.
$$
The surjective homomorphism is an isomorphism. This proves the <basic hereditary algebra path-algebra theorem>:
$$
\boxed{A\cong kQ.}
$$
For each connected component $C$ of the underlying unoriented graph, $e_C=\sum_{i\in C}e_i$ is a <central idempotent>. Paths never connect distinct components, so $kQ=\bigoplus_Ce_CkQ$ as a <direct sum> of <two-sided ideals>. To prove each is a block, let $z$ be any <central idempotent>. Modulo the arrow ideal it has coordinates $c_i\in\{0,1\}$. Commuting $z$ with an arrow $i\to j$ and reducing modulo the square of that ideal gives $c_i=c_j$. Thus these coordinates are constant on each component. If they are zero on a component, the restriction of $z$ there is a nilpotent <idempotent>, hence zero; if they are one, apply that argument to $1-z$. Consequently
$$
\boxed{\text{blocks of }A\ \longleftrightarrow\ \text{connected components of the underlying graph of }Q.}
$$
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