transition function $P_t(k,\ell)$ for a Markov-process with $p(m,m+1)=1$.

Consider the $$Q$$-matrix on $$\mathbb{N}$$ defined by $$q(m,m+1)=-q(m,m)=\rho m$$ and $$q(m,m')=0$$ if $$|m-m'|>1$$, where $$\rho$$ is a constant. Letting $$(P_t)$$ be the transition function of this process, to show is that for $$k\leq \ell$$ $$P_t(k,\ell) = {\ell-1\choose \ell-k}e^{-k\rho t}(1-e^{-\rho t})^{\ell-k}$$ I have already determined that $$p(m,m+1) = 1$$ and each stay at state $$m$$ is $$\sim \exp(\rho m)$$.

Every try based on only calculations resulted in mess. I think we must use that something in this system is binomial, because the final expression looks like that. To come to there, we can think about the reserved process $$(Y_t)$$ of $$(X_t)$$ and transform the system a bit: if $$Y_t$$ is the number of particles at time $$t$$, then if we assume that each of the particles dies exponentially with rate $$\rho$$, then $$(Y_t)$$ is indeed the reversed process of $$(X_t)$$. This is because $$p(m,m-1) = 1$$ for $$(Y_t)$$ and $$p(m,m+1)=1$$ for $$(X_t)$$, and the lengths of periods at all states $$m$$ have the same distribution at both $$(X_t)$$ and $$(Y_t)$$. But I don’t know how to use this further. Bayes rule seems to be essential, but we need to know $$P(X_s = k)/ P(X_{s+t} = \ell)$$ for this.

Is it doable to calculate $$P(X_s = k)/P(X_{s+t} = \ell)$$, or would I yet better just calculate $$P_t(k,\ell)$$ without using the fact that something is binomial?