= Solution
In Kendall notation, the first two letters $M$ mean <Poisson process>[Poisson arrivals] and <exponential distribution>[exponential service times]. An <M M 1 queue> has arrival rate $\lambda$, one server of rate $\mu$, and queue-length transitions
$$
n\to n+1\text{ at rate }\lambda,
\qquad
n\to n-1\text{ at rate }\mu\quad(n\geq1).
$$
It has a <stationary distribution of an M-M-1 queue> exactly when
$$
\rho:=\frac\lambda\mu<1,
$$
and then
$$
\boxed{\pi_n=(1-\rho)\rho^n,
\qquad n\geq0.}
$$
An <M-M-infinity queue> has infinitely many servers, so every customer begins service immediately. Its occupancy transitions are
$$
n\to n+1\text{ at rate }\lambda,
\qquad
n\to n-1\text{ at rate }n\mu.
$$
The <detailed balance for a birth-death process> equations give
$$
\pi_{n+1}=\pi_n\frac{\lambda}{(n+1)\mu}.
$$
Thus for every $\lambda,\mu>0$ the <stationary distribution of an M-M-infinity queue> exists and is
$$
\boxed{\pi_n=e^{-\rho}\frac{\rho^n}{n!},
\qquad \rho=\frac\lambda\mu,}
$$
the <Poisson distribution> of mean $\rho$. The <Burke theorem for an M-M-infinity queue> states that in this stationary regime the departure process is a <Poisson process> of rate $\lambda$; equivalently, time reversal turns departures into arrivals. The past departure process is also independent of the number in the system at the observation time.
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