A directed graph has a cycle iff its depth-first search reveals a back edge

Proof:

(1)

If $(u, v)$ is a back edge, then there is a cycle consisting of this edge together with the path from $v$ to $u$ in the search tree.

(2)

If the graph has a cycle $v_0 \rightarrow v_1 \rightarrow \cdots \rightarrow v_k \rightarrow v_0$, look at the first node on this cycle to be discovered. Suppose it is $v_i$, all the other $v_j$ are reachable from it and will therefore be its descendants in the search tree. In particular, the edge $v_{i-1} \rightarrow v_i$ leads from a node to its ancester and is thus by definition a back edge.