Series with an exponential - Laplace of discrete random variable

Question

I'm trying to compute the Laplace transform of a Poisson random variable with parameter $\lambda$.

$E[e^{-tX}] = e^{-\lambda}\sum_{k=0}^{\infty} e^{-\lambda t} \frac{\lambda^k}{k!}$

I don't know how to continue now.

Answer 1

This should be $$ \mathbb{E}[e^{-tX}] = e^{-\lambda}\sum_{k=0}^{\infty} e^{-{\color{red}kt}} \frac{\lambda^k}{k!} \tag{1}$$ not$$ \mathbb{E}[e^{-tX}] = e^{-\lambda}\sum_{k=0}^{\infty} e^{-{\color{red}\lambda t}} \frac{\lambda^k}{k!}\,.\tag{2} $$ (Can you see why?) Then, with the correct expression (1), you can continue as $$ \mathbb{E}[e^{-tX}] = e^{-\lambda}\sum_{k=0}^{\infty}\frac{(e^{-t} \lambda)^k}{k!} = e^{-\lambda}e^{e^t \lambda} = e^{(e^{-t} -1) \lambda} \tag{3} $$ as desired.