How do I evaluate $\sum_{r = 0}^{n^2} (-1)^r \binom{n^2+n}{r}$? 
$$\sum_{r = 0}^{n^2} (-1)^r \binom{n^2+n}{r}$$

This is what Wolfram says:

Is there any simple way to prove it? Like with counting arguments or manipulating binomial expansions? 
 A: Using Vandermonde's Identity:
$$
\begin{align}
\sum_{r=0}^{n^2}(-1)^r\binom{n^2+n}{r}
&=\sum_{r=0}^{n^2}(-1)^{n^2}\binom{n^2+n}{r}\binom{-1}{n^2-r}\\
&=(-1)^{n^2}\binom{n^2+n-1}{n^2}
\end{align}
$$
A: This is a case of
$$\sum_{r=0}^k(-1)^r\binom mr=(-1)^k\binom{m-1}k$$
which is easily proved by induction on $k$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{r = 0}^{n^{2}}\pars{-1}^{r}{n^{2} + n \choose r} & =
\sum_{r = 0}^{n^{2}}\pars{-1}^{r}{n^{2} + n \choose n^{2} + n - r} =
\sum_{r = 0}^{n^{2}}\pars{-1}^{r}
\bracks{z^{n^{2} + n - r}}\pars{1 + z}^{n^{2} + n}
\\[5mm] & =
\bracks{z^{n^{2} + n}}\pars{1 + z}^{n^{2} + n}\sum_{r = 0}^{n^{2}}\pars{-z}^{r} =
\bracks{z^{n^{2} + n}}\pars{1 + z}^{n^{2} + n}
\,{\pars{-z}^{n^{2} + 1} - 1 \over -z - 1}
\\[5mm] & =
\bracks{z^{n^{2} + n}}\pars{1 + z}^{n^{2} + n - 1} +
\pars{-1}^{n^{2}}\bracks{z^{n - 1}}\pars{1 + z}^{n^{2} + n - 1}
\\[5mm] & =\
\underbrace{{n^{2} + n - 1 \choose n^{2} + n}}_{\ds{=\ 0}}\ +\ \pars{-1}^{n^{2}}{n^{2} + n - 1 \choose n - 1} =
\bbx{\pars{-1}^{n^{2}}{n^{2} + n - 1 \choose n - 1}}
\end{align}
