That's the number of all directed graphs, some of which certainly have cycles.

So it is. 3^(n choose 2) >> n^n stands though.

A lower bound for the number of DAGs can be found by observing that if we drop the directedness of the edges, there are 2^(n choose 2) undirected graphs on a set of n distinguishable vertices, and each of these corresponds to at least 1 DAG. Therefore there are at least that many DAGs, and 2^(n choose 2) is also much larger than n.