Suppose I have an $M/G/1$-queue $Q$. Then I have an embedded Markov chain $Q(D)$ and the following theorem:
If $\rho=\lambda \mathbb{E}[S]<1$, then $Q(D)$ is ergodic with a unique stationary distribution $\pi$, having generating function: $$G(s)=\sum_j \pi_j s^j=(1-\rho)(s-1) \frac{M_S(\lambda(s-1))}{s-M_S(\lambda(s-1))},$$ where $M_S$ is the moment generating function of the service time $S$.
Now I want to apply this theorem to an $M/M/1$-queue. So I know that the service times $S$ are exponential distributed with parameter $\mu$. Therefore the moment generating function of $S$ is: $$M_S(t)=\frac{\mu}{\mu-t}.$$ Now I calculatded the $G(s)$ of the theorem and got: $$G(s)=\frac{1-\rho}{1-s\rho}.$$ Since I know that an $M/M/1$-queue for $\rho<1$ has the geometric distribution: $$\Pr(Q(t)=n) \rightarrow (1-\rho) \rho^n,$$ as $t\rightarrow \infty$ and $\pi$ is the unique stationary distribution.
But the moment generating function of a geometric random variable is given by: $$G(t)=\frac{1-\rho}{1-\rho e^t}.$$ So it isn't exactly the same as $G(s)=\frac{1-\rho}{1-s\rho}$... Can somebody explain me how can I conclude from $G(s)=\frac{1-\rho}{1-s\rho}$ that the queue is gemetric distributed? I appreciate any help! Thank you!!