What is the expected time do i have to wait until n events are completed each distributed in exponential time $\mu$?
I thought that $1/\mu$ is the expectation for the first since the events are coming as poisson processes and their interarrivals times are exponential. So for the $n$ events it is $n/\mu$. Is it correct?
If their distributions are consecutive (i.e. time for the second does not start until the first is done and so on) then you are correct.
If their distributions are concurrent (i.e. time for each of them starts with $t=0$) and they are independent, then the expected time for the first to occur is $\frac{1}{n\mu}$, the expected time between the first and the second occurrence is $\frac{1}{(n-1)\mu}$ and so on up to the expected time between the penultimate and the final occurrence is $\frac{1}{\mu}$ by the memoryless property, making the total time $\dfrac1\mu H_n$ where $H_n$ is the $n$th harmonic number.