Arrival distribution of an M/M/1 queue

Show that the arrivals $$A_n$$ of an M/M/1 queue $$X$$ with initial distribution $$\eta_i := \rho^{i-1}(1-\rho)$$ ($$i \ge 1$$), where $$\rho$$ is the traffic intensity, satisfy $$X_{A_n} \sim \ \eta$$.

I understand that the stationary distribution of the queue is $$\rho^i(1-\rho)$$ for $$i \ge 0$$, and at an arrival the queue cannot be empty, so the queue size should intuitively have the same geometric property but excluding the state 0. How do I show this?

Recall that $$\rho=\lambda/\mu$$, where $$\lambda$$ is the arrival rate and $$\mu$$ the service rate.
For $$j \ge 2$$, $$\mathbb{P}(X_{A_1}=j) = \sum_{i=j-1}^\infty\mathbb{P}(\text{there are exactly} \ n\ \text{departures before the first arrival}) \\ = \sum_{i=j-1}^\infty\rho^{i-1}(1-\rho)\left(\frac{\mu}{\lambda+\mu}\right)^{i-j+1}\left(\frac{\lambda}{\lambda+\mu}\right)\\ =\sum_{k=j-2}^\infty\rho^k\left(\frac{\mu}{\lambda+\mu}\right)^k(1-\rho)\left(\frac{\lambda}{\lambda+\mu}\right)\left(\frac{\mu}{\lambda+\mu}\right)^{-j+2}\\=\sum_{k=j-2}^\infty\left(\frac{\lambda}{\lambda+\mu}\right)^k(1-\rho)\left(\frac{\lambda}{\lambda+\mu}\right)\left(\frac{\lambda+\mu}{\mu}\right)^{j-2}\\=\frac{\left(\frac{\lambda}{\lambda+\mu}\right)^{j-2}}{\frac{\mu}{\lambda+\mu}}(1-\rho)\left(\frac{\lambda}{\lambda+\mu}\right)\left(\frac{\lambda+\mu}{\mu}\right)^{j-2}\\=\left(\frac{\lambda}{\mu}\right)^{j-2}(1-\rho)\left(\frac{\lambda}{\mu}\right)\\=\rho^{j-1}(1-\rho)$$ as required. For $$j=0$$, note that the queue is non-empty when a customer has just arrived so $$\mathbb{P}(X_{A_1}=0)=0$$. For $$j=1$$, the distribution must sum to 1, yielding $$\mathbb{P}(X_{A_1}=1)=1-\rho$$.
Hence $$X_{A_1} \sim \eta$$. By the strong Markov property (for the M/M/1 queue), $$(X_{A_n})_{n\ge0}$$ is a discrete-time Markov chain which starts in $$\eta$$ and has its first arrival given by $$\eta$$, so the Markov property (for the chain), $$\forall n\ge0: X_{A_n} \sim \eta$$.
$$\square$$