Since $D$ is strongly connected it has an ear decomposition $D_1, \ldots, D_n \subset D$ where the $D_i$ are directed paths.
If $D_n$ has length greater than 1, then it has an internal vertex $v$. Since $D_1, \ldots, D_n$ was an ear decomposition, this vertex cannot appear in any $D_i$ for $i < n$. Thus $v$ has indegree and outdegree of exactly $1$.
If $D_n = (v_1, v_2) $ has length exactly 1, then $v_1$ and $v_2$ already appear in some $D_i$ and $D_j$ for $i, j < n$. This means that removing the edge $(v_1, v_2)$ from $D$ leaves a strongly connected graph, which contradicts the fact that $D$ was minimally strongly connected.