An operator \(T\) on a separable complex Hilbert space \(\mathcal H\) space is said to be hypercyclic if there is a vector \(x\) such that the orbit \(\{T^nx: n=0,1,\ldots\}\) is dense in \(\mathcal H\). An operator is said to be supercyclic if there is a vector \(x\) such that the scalar multiples of the elements in the orbit are dense in \(\mathcal H\). The operator \(T\) is said to be multi-hypercyclic if there are vectors \(x_1,\ldots ,x_n\) in \(\mathcal H\) such that the union of their orbits under \(T\) is dense in \(\mathcal H\). Multi-supercyclicity is defined in the obvious way. The author proves the following important conjecture of \textit{Domingo A. Herrero} [J. Oper. Theory 28, No. 1, 93-103 (1992; Zbl 0806.47020)]: every multi-hypercyclic operator is hypercyclic. He also proves the corresponding result for supercyclic operators. The proof, which is very interesting, uses some fine areguments related to previous work by \textit{Sh. I. Ansari} [J. Funct. Anal. 128, No. 2, 374-383 (1995; Zbl 0853.47013)]. The results are obtained in the general context of complex and real locally convex spaces. The author finishes with a very interesting question: Assume that the closure of the orbit of a vector \(x\) under an operator \(T\) contains an open set. is the operator hypercyclic? Although a negative answered is conjectured, \textit{Bourdon} and \textit{Feldman} [``Somewhere dense orbits are everywhere dense'', preprint] have recently proved that operators with somewhere dense orbits are also hypercyclic, which, obviously, also implies that multi-hypercyclic operators are hypercyclic.