I studied this 10+ years ago. But I just saw a very concise derivation from stack exchange

Let \(N_t\) be a Poisson counter process, with parameter \(\lambda>0\). Let there be another stochastic process \(X_t = f(t,N_t)\). Then, by Taylor series expansion,

\[\begin{align} dX_t &= \frac{\partial X_t}{\partial t}dt + \frac{\partial X_t}{\partial N_t}dN_t \\ &\quad + \frac{1}{2!}\left( \frac{\partial^2 X_t}{\partial N_t^2}(dN_t)^2 + 2\frac{\partial^2 X_t}{\partial N_t\partial t}dN_tdt + \frac{\partial^2 X_t}{\partial t^2}(dt)^2 \right) \\ &\quad + \frac{1}{3!}\left( \frac{\partial^3 X_t}{\partial N_t^3}(dN_t)^3 + 3\frac{\partial^3 X_t}{\partial N_t^2\partial t}(dN_t)^2dt + 3\frac{\partial^3 X_t}{\partial N_t\partial t^2}dN_t(dt)^2 + \frac{\partial^3 X_t}{\partial t^3}(dt)^3 \right) + \cdots \\ &= \frac{\partial X_t}{\partial t}dt + \left(\frac{\partial X_t}{\partial N_t} + \frac{1}{2!}\frac{\partial^2 X_t}{\partial N_t^2} + \frac{1}{3!}\frac{\partial^3 X_t}{\partial N_t^3} + \cdots \right)dN_t \end{align}\]

The second equality is due to the fact that \(dN_tdt = 0\), and \(dN_t = (dN_t)^2 = (dN_t)^3 = \cdots = (dN_t)^n\) since \(dN_t \in \{0, 1\}\).

Therefore, we have

\[dX_t = \frac{\partial X_t}{\partial t}dt + \left(\sum_{n=1}^{\infty}\frac{1}{n!}\frac{\partial^n X_t}{\partial N_t^n}\right)dN_t\]

For example, if

\[X_t = \left(\frac{\eta}{\lambda}\right)^{N_t} e^{(\lambda-\eta)t}\]

then

\[\begin{align} \frac{\partial X_t}{\partial t} &= (\lambda-\eta)X_t \\ \frac{\partial^n X_t}{\partial N_t^n} &= \left[\log\left(\frac{\eta}{\lambda}\right)\right]^n X_t \\ \therefore\ dX_t &= \frac{\partial X_t}{\partial t}dt + \left(\sum_{n=1}^{\infty}\frac{1}{n!}\frac{\partial^n X_t}{\partial N_t^n}\right)dN_t \\ &= (\lambda-\eta)X_t dt + \left(\sum_{n=1}^{\infty}\frac{1}{n!} \left[\log\left(\frac{\eta}{\lambda}\right)\right]^n X_t \right) dN_t \\ &= (\lambda-\eta)X_t dt + \left(\exp\left(\log\left(\frac{\eta}{\lambda}\right)\right)-1\right) X_t dN_t \\ &= (\lambda-\eta)X_t dt + \left(\frac{\eta}{\lambda}-1\right) X_t dN_t \end{align}\]