Expression $\lambda e^{-\lambda} \sum_{k \ge 1} \frac{1}{(k - 1)!} \lambda^{k - 1}$ in expectation of Poisson distribution proof I am trying to follow this expectation of Poisson distribution proof:

From the definition of expectation:
$$E(X) = \sum_{x \ \in \ \text{Img}(x)} x P(X = x)$$
By definition of Poisson distribution:
$$E(X) = \sum_{k \ge 0} k \dfrac{1}{k!} \lambda^k e^{-\lambda}$$
Then:
$$\begin{align} E(X) &= \lambda e^{-\lambda} \sum_{k \ge 1} \dfrac{1}{(k - 1)!} \lambda^{k - 1} \ \ \ \ \text{as the $k = 0$ term vanishes} \\ &= \lambda e^{-\lambda} \sum_{j \ge 0} \dfrac{\lambda^j}{j!} \ \ \ \ \text{putting $j = k - 1$} \\ &= \lambda e^{-\lambda} e^{\lambda} \ \ \ \ \text{Taylor Series Expansion for Exponential Function} \\ &= \lambda \end{align}$$

I am confused by how the author proceeds from $\sum\limits_{k \ge 0} k \dfrac{1}{k!} \lambda^k e^{-\lambda}$. Specifically, since the term becomes $0$ for $k = 0$, shouldn't we then have $\sum\limits_{k \ge 1} k \dfrac{1}{k!} \lambda^k e^{-\lambda}$ for the proceeding terms? So I don't understand how the $k = 0$ term vanishing means that the expression must be $\lambda e^{-\lambda} \sum\limits_{k \ge 1} \dfrac{1}{(k - 1)!} \lambda^{k - 1}$.
I would greatly appreciate it if people would please take the time to clarify this.
 A: Your expression is the same as the one that they write down. Note that since $k / k! = 1 / (k-1)!$,
$$\sum_{k \geq 1} k \frac{1}{k!} \lambda^k e^{-\lambda} = \sum_{k \geq 1} \frac{1}{(k-1)!} \lambda^k e^{-\lambda}$$
But since the sum only depends on $k$, we can take out any factors involving only $\lambda$. Specifically,
$$\sum_{k \geq 1} \frac{1}{(k-1)!} \lambda^k e^{-\lambda} = \sum_{k \geq 1} \frac{1}{(k-1)!} \lambda^{k-1} (\lambda e^{-\lambda}) = \lambda e^{-\lambda} \sum_{k \geq 1} \frac{1}{(k-1)!} \lambda^{k-1} $$
But now, the summand only depends on $k - 1$, which suggests that we should define a new summation index (call it $j$) that equals $k - 1$. Since summing $k$ from $1$ to $\infty$ is equivalent from summing $j = k-1$ from $0$ to $\infty$, it follows that
$$\lambda e^{-\lambda} \sum_{k \geq 1} \frac{1}{(k-1)!} \lambda^{k-1} = \lambda e^{-\lambda} \sum_{j \geq 0} \frac{1}{j!} \lambda^j$$
But the sum is just the Taylor series of $e^{\lambda}$, so our expression simplifies to $\boxed{\lambda}$, as desired.
