Explanation step in Wikipedia proof of the expectation of Binomial RV from first principles

Question

My question is very simple: why does the index in the third step go from $0$ to $1$? No change of variable seems to happen, the only thing that happens is a rewriting and cancellation of $k$.. can you help me understand it? $$ \begin{align} \mu &= \sum_{k=0}^nk\binom nkp^k(1-p)^{n-k}\\ &=np\sum_{k=0}^nk\frac{(n-1)!}{(n-k)!k!}p^{k-1}(1-p)^{(n-1)-(k-1)}\\ &=np\sum_{k=1}^n\frac{(n-1)!}{((n-1)-(k-1))!(k-1)!}p^{k-1}(1-p)^{(n-1)-(k-1)}\\ &=np\sum_{k=1}^n\binom{n-1}{k-1}p^{k-1}(1-p)^{(n-1)-(k-1)}\\ &=np\sum_{l=0}^{n-1}\binom{n-1}lp^l(1-p)^{(n-1)-l}\\ &=np\sum_{l=0}^{m}\binom{m}lp^l(1-p)^{m-l}\\ &=np(p+(1-p))^m\\ &=np \end{align} $$

Answer 1

Because the term with $k=0$ vanishes.

If $a_k\in\mathbb{R}$ and $a_0=0$ then $$\begin{align*} \sum_{k=0}^na_k &= a_0+a_1+a_2+\cdots+a_n \\ &= 0+(a_1+a_2+\cdots+a_n) \\ &=a_1+a_2+\cdots+a_n \\ &=\sum_{k=1}^na_k. \end{align*}$$

A clearer proof would have begin with this issue first: $$\begin{align*} \mu &= \sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k} \\ &= \sum_{k=1}^nk\binom{n}{k}p^k(1-p)^{n-k} \\ &= \cdots \end{align*} $$

Answer 2

$$\frac{k}{k!} = \begin{cases} 0 & k=0 \\ \frac{1}{(k-1)!} & k>0 \end{cases}.$$

Thus they have removed the $k=0$ term and simplified the remaining terms.