How does this vandermonde identity proof works? Recently I have taken a look at this vandermonde identity proof Inductive Proof for Vandermonde's Identity?.
\begin{align*}
  \binom{m + (n+1)}r &= \binom{m+n}r + \binom{m+n}{r-1}\\
       &= \sum_{k=0}^r \binom mk\binom n{r-k} + \sum_{k=0}^{r-1} \binom mk\binom{n}{r-1-k}\\
       &= \binom mr + \sum_{k=0}^{r-1} \binom mk\biggl(\binom n{r-k} + \binom n{r-1-k}\biggr)\\
       &= \binom mr\binom{n+1}0 + \sum_{k=0}^{r-1} \binom mk\binom{n+1}{r-k}\\
       &= \sum_{k=0}^r \binom mk \binom{n+1}{r-k}
\end{align*}
I am confused mostly at this part
\begin{align*}
  &= \sum_{k=0}^r \binom mk\binom n{r-k} + \sum_{k=0}^{r-1} \binom mk\binom{n}{r-1-k}\\
  &= \binom mr + \sum_{k=0}^{r-1} \binom mk\biggl(\binom n{r-k} + \binom n{r-1-k}\biggr)\\
\end{align*}
How did this
\begin{align*}
&= \sum_{k=0}^r \binom mk\binom n{r-k} 
\end{align*}
turn into
\begin{align*}
 &= \binom mr
\end{align*}
which identity did this person use? I am such that it could magically become C(m,r)?
I am also confused on how
\begin{align*}
\binom{n+1}0 
\end{align*}
appears in here
\begin{align*}
&= \binom mr\binom{n+1}0 + \sum_{k=0}^{r-1} \binom mk\binom{n+1}{r-k}\\
\end{align*}
If someone could help that would be very helpful, I didn't comment on the post due to it being 8 years, posted already. If someone could help and point where I should study that would be very helpful. Thank you!

Do you mean this step?
$\displaystyle\sum_{k=0}^r {m \choose k}{n \choose r-k} = {m \choose r} + \sum_{k=0}^{r-1}{m\choose k}{n\choose r-k}$
Also for your second question, as $0! \equiv 1$ we'll have ${n+1\choose 0} = 1$.
