= Solution
Let $A_{n+1}$ be the number of arrivals during the next customer's service. If $Q_n>0$, that service begins immediately and one of the waiting customers is served; if $Q_n=0$, the next arrival initiates the service, and it is not counted among arrivals during its own service. Thus
$$
\boxed{Q_{n+1}=(Q_n-1)^++A_{n+1}.}
$$
Conditional on a fresh service time $S=t$, $A_{n+1}$ is Poisson with mean $\lambda t$. Independence of service times and the independent-increment/strong Markov property of the Poisson arrival process make these mixed counts identically distributed and independent of the previous departure history. Hence the conditional law of $Q_{n+1}$ depends only on $Q_n$, proving the <Markov property> of the departure-embedded queue.
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