Proving $\binom {n-1}{r-1}=\sum_{k=0}^r(-1)^k\binom r k \binom{n+r-k-1}{r-k-1}$ 
Prove the identity: $\displaystyle\binom {n-1}{r-1}=\sum_{k=0}^r(-1)^k\binom r k \binom{n+r-k-1}{r-k-1}$

It looks a bit similar to the "no gets their own hat back" problem or inclusion exclusion or non distinct balls in bins. 
Trying to find a combinatorial solution seems like impossible because of the alternating sum (how can we explain inclusion exclusion?).
Trying to expand the RHS doesn't help nor using any of the simple identities I know of (like Pascal's).
Any hints or directions please?
Note: no integrals, no generating functions nor use of other identities without proving them.
Edit: I think I got it: 
LHS: 
n non distinct balls to r bins such that every bin has at least one ball, spread 1 ball to each bin, we're left with n-r balls to r bins.
RHS: 
General case: $\binom {n+r-1}{r-1}$ 
complement: at least one bin is empty; 1 bin is empty, choose that bin $\binom r 1$ and spread the balls: $\binom{n+r-1-1}{r-1-1}$, do this up to r empty bins. 
Since we have many over counting, we'll apply the inclusion exclusion principle and we got what we desired. 
 A: $$\begin{align}
\sum_{k=0}^r(-1)^k\binom rk\binom{n+r-k-1}{r-k-1}
&=\sum_{k=0}^r(-1)^k\binom rk\binom{-n-1}{r-k-1}(-1)^{r-k-1}&&(1)\\
&=(-1)^{r-1}\sum_{k=0}^r\binom rk\binom{-n-1}{r-1-k}\\
&=(_1)^{r-1}\binom{r-n-1}{r-1}&&(2)\\
&=(-1)^{r-1}\binom{n-1}{r-1}(-1)^{r-1}&&(3)\\
&=\binom{n-1}{r-1}\qquad\blacksquare
\end{align}$$

(1): using Upper Negation
(2): using Vandermonde Identity
(3): using Upper Negation
__
$\color{gray}{\text{Proof of Upper Negation}}$
$$\color{gray}{\begin{align}\\
\binom ab&=\frac{a^{\underline{b}}}{b!}
=\frac{a(a-1)(a-2)\cdots(a-b+1)}{b!}\\
&=(-1)^b \frac{[-a][-(a-1)][-(a-2)]\cdots[-(a-b+1)]}{b!}\\
&=(-1)^b \frac{(b-1-a)\cdots(2-a)(1-a)(-a)}{b!}\\
&=(-1)^b \binom{b-a-1}b\end{align}}$$
$\color{gray}{\text{Proof of Vandermonde Identity}}$
$$\color{gray}{\begin{align}\\
\sum_{i=0}^a\binom ai x^i\sum_{j=0}^b\binom bjx^j
&=(1+x)^a(1+x)^b=(1+x)^{a+b}\\
[x^n]:
\sum_{i+j=n}\binom ai\binom bj
&=\underbrace{\sum_{i=0}^n \binom ai\binom b{n-i}=\binom{a+b}n}_{\text{Vandermonde Identity}}
\end{align}}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 0}^{r}\pars{-1}^{k}{r \choose k}{n + r - k - 1 \choose r - k - 1}} =
\sum_{k = 0}^{r}\pars{-1}^{k}{r \choose k}{-n - 1 \choose r - k - 1}
\pars{-1}^{r - k - 1}
\\[5mm] = &\
\pars{-1}^{r + 1}\sum_{k = 0}^{r}{r \choose k}
\bracks{z^{r - k - 1}}\pars{1 + z}^{-n - 1} =
\pars{-1}^{r + 1}\bracks{z^{r - 1}}\pars{1 + z}^{-n - 1}
\sum_{k = 0}^{r}{r \choose k}z^{k}
\\[5mm] = &\
\pars{-1}^{r + 1}\bracks{z^{r - 1}}\pars{1 + z}^{-n - 1 + r} =
\pars{-1}^{r + 1}{-n - 1 + r \choose r - 1}
\\[5mm] = &\
\pars{-1}^{r + 1}{n - 1 \choose r - 1}\pars{-1}^{r - 1} =
\bbx{\large{n - 1 \choose r - 1}} \\ &
\end{align}
A: We seek to evaluate
$$\sum_{k=0}^r (-1)^k {r\choose k} {n+r-k-1\choose r-k-1}
= \sum_{k=0}^r (-1)^k {r\choose k} {n+r-k-1\choose n}.$$
Note that the second binomial coefficient is zero when $k=r$
because $(n-1)^\underline{n} = 0.$ Continuing we find
$$[z^n] (1+z)^{n+r-1}
\sum_{k=0}^r (-1)^k {r\choose k} (1+z)^{-k} 
\\ = [z^n] (1+z)^{n+r-1}
\left(1-\frac{1}{1+z}\right)^r
= [z^n] (1+z)^{n+r-1}
\frac{z^r}{(1+z)^r}
\\ = [z^{n-r}] (1+z)^{n-1}
= {n-1\choose n-r} = {n-1\choose r-1}.$$
