There is an edge emanating from any vertex of $G$ because otherwise the corresponding row (or column, according to our convention for forming the adjacency matrix) would be zero. Label the vertices as $v_1,\dots,v_n$. According to the observation just made, for any $i$, we can pick an index $f(i)$ such that there is a directed edge from $v_i$ to $v_{f(i)}$—thus a function 
$f:\{1,\dots,n\}\rightarrow\{1,\dots,n\}$. There clearly exists an index $k$ and a positive integer $m$ such that $f^{\circ m}(k)=k$ and $k,f(k),\dots,f^{\circ (m-1)}(k)$ are pairwise distinct. Now $v_k,v_{f(k)},\dots,v_{f^{\circ (m-1)}(k)}$ is a directed cycle of $G$.