Replacing kernel for a Poisson distribution in an equation In Example 4.4.2 on page 163 of Statistical Inference, the following is implied:
$\sum_{t=0}^{\infty}\frac{((1 - p)\lambda)^t}{t!} = e^{(1-p)\lambda}$
with a note that says "[the] sum is a kernel for a Poisson distribution". I understand the Poisson distribution, but I don't fully grasp the kernel concept yet. I tried searching for this identity, but couldn't find it. Can someone help me understand this, or provide  references that I can use for background information?
Update: I initially had the incorrect equation above. I corrected it after Anand's comments.
 A: I checked the book and there is a typo in your question. It should be 
\begin{align*} 
\sum_{t=0}^{\infty} \frac{((1-p) \lambda)^t}{t!} & = e^{(1-p) \lambda}
\end{align*}
The above formula basically follows from the definition of an exponential function
\begin{align*} 
\sum_{t=0}^{\infty} \frac{x^t}{t!} & = e^{x}
\end{align*}
and once you substitute $x = (1-p) \lambda$ you get your desired result.
The above result can be derived using the Poisson probability mass function (pmf) which the authors refer to  "[the] sum is a kernel for a Poisson distribution". This follows from the fact the pmf should sum to $1$. 
\begin{align*} 
\sum_{x} \Pr(X = x) &= 1
\end{align*}
The pmf of Poisson is given by 
\begin{align*} 
\Pr(X = x) &= \frac{e^{-\lambda} \lambda^{x}}{x!}
\end{align*}
Using the above two equations, we get 
\begin{align*}
\sum_{x} \Pr(X = x) &= \sum_{x} \frac{e^{-\lambda} \lambda^{x}}{x!} = 1
\end{align*}
Multiply both sides by $e^{\lambda}$ we get 
\begin{align*}
\sum_{x} \frac{\lambda^{x}}{x!} = e^{\lambda}
\end{align*}
