How to calculate $E[e^{tX}\mid\Lambda=\lambda]$ given $P\{E[X\mid\Lambda=\lambda]=k\}=\frac{e^{-\lambda}\lambda^k}{k!}$?

Question

Let $X,\Lambda$ be random variables such that $M_\Lambda(t)<\infty$ for $t\in]-1,1[$ and $P\{E[X\mid\Lambda=\lambda]=k\}=\frac{e^{-\lambda}\lambda^k}{k!}$ for $k\in\mathbb{N}_0$ (so Poisson$(\lambda)$). Then how to calculate $E[e^{tX}\mid\Lambda=\lambda]$? Intuitively I would say that since $Y:=E[X\mid\Lambda=\lambda]$ is discrete with $P\{Y\in\mathbb{N}_0\}=1$ then also $P(X\in\mathbb{N}_0)=1$ and thus $$ E[e^{tX}\mid\Lambda=\lambda]=\sum_{k=0}^\infty e^{tk}\frac{e^{-\lambda}\lambda^k}{k!}=\exp\left({\lambda\left(e^t-1\right)}\right). $$ However I cannot formalize it. Is this calculation correct? And if so, how to do it more formally?