Exponential Kth Moment Derivation

I have essentially a mathematical question, relating to deriving the formula for the kth moment of an exponential. I can't seem to work out how we get from the 2nd line to the 3rd line; i.e. the fraction to the expansion: \begin{align}%\label{} \nonumber M_X(s)&=\frac{\lambda}{\lambda-s}\\ \nonumber &=\frac{1}{1-\frac{s}{\lambda}}\\ \nonumber &=\sum_{k=0}^{\infty} \left(\frac{s}{\lambda}\right)^k, \hspace{10pt} \textrm{for }\left|\frac{s}{\lambda}\right|<1\\ \nonumber &=\sum_{k=0}^{\infty} \frac{k!}{\lambda^k} \frac{s^k}{k!}. \end{align} My maths is quite rusty so this is probably something simple but I would appreciate if someone could take a minute to help me out. Cheers!

• $\sum_{n=0}^\infty x^n$ is an infinite GP series which equals $1/(1-x)$ provided $|x|<1$. Pretty standard result. This answer on math.se is relevant. – StubbornAtom Jun 8 '18 at 15:59
• @StubbornAtom - may as well re-post as an answer and get the points, since you did answer it! – jbowman Jun 8 '18 at 16:21
• This is an application of the Binomial Theorem. Even if you don't know how the formula was derived, you can verify it by multiplying the sum by $1-s/\lambda$, verifying (since all terms but the first cancel out) that it produces $1.$ Indeed, I recall learning this from a high school algebra textbook that began with the sum--which is a geometric series--and performed this multiplication to derive the equality. – whuber Jun 8 '18 at 16:44