$(N_t,t\ge 0)$ is a poisson process with rate $\lambda$
Using the Kolmogorov backward equation, find the PGF:
$$G_i(z,t)=\Bbb E[z^{N_t}|N_o=i]$$
The answer is $G_i(z,t)=e^{\lambda t(z-1)}$ but I am struggling to get to here. What I have done so far is:
$$G_i(z,t)=\sum_{k\ge 0}z^k\Bbb P(N_t=k|N_0=i)=\sum_{k\ge i}z^kp_{i,k}(t)=\sum_{n=0}^\infty z^{i+n}p_{i,i+n}(t)$$
The Kolmogorov backward equations are given as:
$$\begin{align} \frac{d}{dt}p_{i,i}(t)&=-\lambda p_{i,i}(t) \\ \frac{d}{dt}p_{i,i+n}(t)&=p_{i,i+n-1}(t)\lambda-p_{i,i+n}(t)\lambda \end{align}$$
Using this information we get that:
$$\begin{align} \frac{d}{dt}G_i(z,t)&=\sum_{n=0}^{\infty}z^{i+n}\frac{d}{dt}p_{i,i+n}(t) \\ &= -\lambda z^0p_{i,i}(t)+\lambda \sum_{n=1}^\infty z^{i+n}(p_{i,i+n-1}(t)-p_{i,i+n}(t)) \\ &= \lambda(z-1)G_i(z,t) \end{align}$$
From here it is clear to infer the answer but I am really stuck on how to get the last line of working from the penultimate line. It would be much appreciated if somebody could show me the steps in between of how this happens as I have been failing to get the intuition behind this for some time.