Prove that $\sum_{k=0}^{n-r} \binom{r+k}{r} = \binom{n+1}{r+1}$ In my combinatorics book I have the following identity:
$$\binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} +  ... + \binom{n}{r} = \binom{n+1}{r+1}$$
More succinctly:
$$\sum_{k=0}^{n-r} \binom{r+k}{r} = \binom{n+1}{r+1}$$
I'm trying to reason about why this is the case. My textbook actually "proves" this identity (although I'm not convinced).

We break the ways to pick r + 1 members of a committee from n + 1
  people into cases depending on who is the last person chosen: the (r +
  1)st, the (r + 2)nd, . . . , the (n + 1)st. If the (r + k + 1)st
  person is the last chosen, then there are C(r + k, r) ways to pick
  the first r members of the committee. [The identity] now follows.
  Q.E.D.

So... what are they trying to tell me? I'm guessing we need to partition the committee selection into multiple parts. But what do I partition the committee selection into, and why?
Of course, an algebraic proof is trivial for small examples, but is not practicable for the general case. Can anybody reword my book's committee-selection argument or offer a clearer proof?
 A: Actually, an algebraic proof is possible, here's an example of one. Fix an integer $r$. We'll make induction over $n$. Of course, we need to have $n \geq r$, so our base case will be $n=r$. In that case, we have to prove
$$\binom{r}{r}=\binom{n+1}{r+1}$$
And it's obvious, because $n=r$ implies $\binom{n+1}{r+1}=\binom{r+1}{r+1}=1$. For the induction step, let's assume 
$$\binom{r}{r}+\binom{r+1}{r}+\dots+\binom{n}{r}=\binom{n+1}{r+1} $$
Adding $\binom{n+1}{r}$ on both sides, we have
$$\binom{r}{r}+\binom{r+1}{r}+\dots+\binom{n}{r}+\binom{n+1}{r}=\binom{n+1}{r+1}+\binom{n+1}{r} $$
But $\binom{n+1}{r}+\binom{n+1}{r+1}=\binom{n+2}{r+1}$, which gives us
$$ \binom{r}{r}+\binom{r+1}{r}+\dots+\binom{n}{r}+\binom{n+1}{r} = \binom{n+2}{r+1}=\binom{(n+1)+1}{r+1} $$
So the identity is valid for $n+1$ too. By the Principle of Induction, we have
$$ \binom{r}{r}+\binom{r+1}{r}+\dots+\binom{n}{r}=\binom{n+1}{r+1} $$
for each $r$ integer and $n\geq r$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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An analytical approach:

\begin{align}
\sum_{k = 0}^{n - r}{r + k \choose r} & =
\sum_{k = 0}^{n - r}\oint_{\verts{z} = 1}{\pars{1 + z}^{r + k} \over z^{r + 1}}
\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z} = 1}{\pars{1 + z}^{r} \over z^{r + 1}}
\sum_{k = 0}^{n - r}\pars{1 + z}^{k}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z} = 1}{\pars{1 + z}^{r} \over z^{r + 1}}
{\pars{1 + z}^{n - r + 1} - 1 \over \pars{1 + z} - 1}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z} = 1}{\pars{1 + z}^{n + 1} \over z^{r + 2}}
\,{\dd z \over 2\pi\ic} -
\oint_{\verts{z} = 1}{\pars{1 + z}^{r} \over z^{r + 2}}
\,{\dd z \over 2\pi\ic} =
{n + 1 \choose r + 1} - {r \choose r + 1}
\\[5mm] & = \bbx{\ds{n + 1 \choose r + 1}}
\end{align}
