\displaystyle \sum_{l = 0}^{\infty} \frac{\mu^{k-l} \lambda^l}{(k-l)! \, l!} = \frac{1}{k!} \sum_{l = 0}^{\infty}\binom{k}{l} \mu^{k-l} \lambda^l = \frac{(\mu + \lambda)^k}{k!}