Is the shortest path in Bellman-Ford a simple path?

Question

I think that if there are no negative cycles in a graph, the shortest path from s to t must be a simple one. Am I correct? the way i see it is, that if the same vertex v is repeated in a shortest path, then there's a cycle in that shortest path, and since there's a cycle in the shortest path that means that walking in that cycle decreased our weight (otherwise we would've skipped it) and that means it must be of negative weight. Does that make sense?