Let $X$ be a random variable with distribution
$$P(X=x)=\binom{x-1}{k-1}p^k(1-p)^{x-k}\quad \text{ if } x=k, k+1, \cdots$$
How can I compute the moment generating function $\mathbb E(e^{tX})$?
I know that $$\mathbb E(e^{tX})=\sum_{x=k}^\infty \binom{x-1}{k-1}e^{tx}p^k(1-p)^{x-k}=\sum_{x=k}^\infty \binom{x-1}{k-1}(e^{t}(1-p))^xp^k(1-p)^{-k}$$
But I don't know how to go on.
Just rearrange slightly with a change of bound variable: $$\mathsf E(\mathrm e^{tX})=\dfrac{p^k}{(1-p)^{k}}\sum_{x=k}^\infty\dbinom{x-1}{k-1}((1-p)\mathrm e^{t})^x\\={p^k\mathrm e^{tk}}\sum_{y=0}^\infty\dbinom{y+k-1}{k-1}((1-p)\mathrm e^{t})^{y}\\={p^k\mathrm e^{tk}}\sum_{y=0}^\infty\dbinom{y+k-1}{y}((1-p)\mathrm e^{t})^{y}$$
Then use: $$\sum_{y=0}^\infty\binom{y+n}{y} r^y = (1-r)^{-(n+1)}$$
