Here is the argument given in section 1.3 of Gregory F. Lawler's Introduction to Stochastic Processes. It treats stochastic matrices $P$,
but I think the argument applies to general non-negative matrices.
For each state $i$ define $J_i=\{n: P^n(i,i)>0\}$. This is a semigroup and since
we have assumed that $P$ is aperiodic, we have $\gcd(J_i)=1$ and it follows that
$J_i$ contains all sufficiently large integers.
That is, there is some integer $M(i)$ so that for all $n\geq M(i)$ we have
$P^n(i,i)>0$.
Since $P$ is irreducible, there exists some $m(i,j)$ such that $P^{m(i,j)}(i,j)>0$.
Hence for $n\geq M(i)$,
$$P^{n+m(i,j)}(i,j)\geq P^n(i,i)\,P^{m(i,j)}(i,j)>0.$$
Let $M$ be the maximum value of $M(i)+m(i,j)$ over all pairs $(i,j)$. Then
for $n\geq M$, $P^n(i,j)>0$ for all states $i,j$.
Essentially the same argument is found in section 1.3 of
Markov Chains and Mixing Times by Levin, Peres, and Wilmer. So it looks like probabilists have not found a better proof.
I've come to a solution somewhat simpler.
Let D be a strongly connected digraph. Let $C = v_1 v_2 ... v_m$, $v_m = v_1$, be an odd cycle in the underlying graph. For each i, from 1 to m-1, let $W_i$ be a minimal walk from $v_i$ to $v_{i+1}$. If both vertices are connected by an edge $e_i = v_iv_{i+1}$, then $W_i = v_ie_iv_{i+1}$. Else, if the edge is $v_{i+1}v_i$, then suppose $W_i$ is an odd length walk (otherwise, we'd have an odd length cycle and be done by now). Since m-1 is odd, we have constructed an odd number of odd length walks. Concatenating them as $W_0 . W_1 . ... W_{m-1}$, we have a closed odd walk. Since every closed odd walk has a closed odd cycle, we're done.
Best Answer
You can interpret $(A^n)_{i,j}$ as the number of paths with length $n$ from vertex $i$ to vertex $j$.
Since we have $A^k>0$, we also have, for $n\in\mathbb N$, that $A^{nk}>0$.
Therefore, we can find paths of arbitrary length in the graph, so it must contain a cycle.