Context: Introduction to SDEs, by Evans, page 72, first line.
The author first asserts
$$\frac{d^n}{d\lambda ^n}(e^{\lambda x - \frac{\lambda^2 t}{2}})|_{\lambda=0} = n! h_n(x,t)$$
and then writes
Hence $$e^{\lambda x - \frac{\lambda^2 t}{2}} = \sum\limits_{n=0}^{\infty}\lambda^n h_n(x,t).$$
Also, $h_n(x,t)$ (where $n$ is a non-negative integer) is defined as follows: $$h_n(x,t):= \frac{(-t)^n}{n!}e^{x^2/2t}\frac{d^n}{dx^n}(e^{-x^2/2t}).$$ My question is: how does the equation in the yellow box follow from the previous one?
It's a Taylor series in $\lambda$.
$$\begin{align*}e^{\lambda x - \frac{\lambda^2 t}{2}} &= \sum_{n=0}^{\infty}\lambda^n\frac{f^n(0)}{n!}\\&=\sum_{n=0}^{\infty}\lambda^n\frac{n!h_n(x,t)}{n!}\\&=\sum_{n=0}^{\infty}\lambda^nh_n(x,t)\end{align*}$$