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Using the Power Series expansion for the natural exponential,
\begin{align*} \sum_{x=0}^{\infty} f(x) & = \sum_{x=0}^{\infty} \frac{(\lambda T)^x}{x!} e^{-\lambda T} \\ & = e^{-\lambda T} \sum_{x=0}^{\infty} \frac{(\lambda T)^x}{x!} \\ & = e^{-\lambda T} e^{\lambda T} \\ & = 1 \end{align*}
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