Looking at the number of successes in a unit interval in the poisson process, I would imagine knowing we have at least one success simply shifts the pdf function for number of succes in a time interval of a poisson process to the right by one (since success at any instance of the poisson process is independent), giving us $ p_x(k) = e^{-\lambda}\frac{\lambda^{k-1}}{(k-1)!}\forall k \geq 1$. If we try to calculate it with the usual definition of conditional probability, however, $$P(X = k | X \geq 1) = \frac{P(X=k)}{P(X \geq 1)}=\frac{e^{-\lambda}\frac{\lambda^k}{k!}}{\sum_{t=1}^{\infty}e^{-\lambda}\frac{\lambda^t}{t!}}=\frac{\lambda^k}{k!(-1+e^\lambda)} \forall k \geq 1$$
I checked in Wolfram and these two expressions don't equal each other, so I was wondering where I went wrong (probably either with my thought process on the Poisson process, with my calculations on conditional probability, or with some reason why these two calculation are looking at different probability models / have different assumptions)
Thank you!