Some curious binomial coefficient identities I was playing around with some polynomials involving binomial coefficients, and inadvertently proved the following two identities:
(1) For all $p, q \in \mathbb{N}$ and $i \in \left[ 0, \min\{p, q\} \right]$:
$$
\begin{pmatrix}p \\ i\end{pmatrix} = \sum_{j=0}^i (-1)^j \begin{pmatrix}q \\ j\end{pmatrix} \begin{pmatrix}p + q - j \\ i - j\end{pmatrix} \text{.}
$$
(2) For all $q \in \mathbb{N}_{\ge 1}$ and $i \in [0, q]$:
$$
\frac{q - i}{q} = \sum_{j=0}^i (-1)^j \begin{pmatrix}i \\ j\end{pmatrix} \begin{pmatrix}2q - 1 - j \\ i - j\end{pmatrix} \begin{pmatrix}q - j \\ i - j\end{pmatrix}^{-1}  \text{.}
$$
Can either of these identities be proven in any trivial way (e.g., by reduction to known identities)?

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$\ds{\sum_{j = 0}^{i}\pars{-1}^{\,j}{q \choose j}{p + q - j \choose i - j} =
{p \choose i}.\qquad p, q \in \mathbb{N}\,;\quad i \in \bracks{0,\min\braces{p,q}}}$.

From the above conditions it's clear, because $\ds{j \leq i}$, that
$\ds{p + q - j \geq 0}$. Then, $\ds{{p + q - j \choose i - j}_{j\ >\ i} = 0}$ such that $\color{#f00}{the\ above\ sum\ can\ be\ extended}$ to all numbers $\ds{\in \mathbb{N}}$:
\begin{align}
&\sum_{j = 0}^{i}\pars{-1}^{\,j}{q \choose j}{p + q - j \choose i - j} =
\sum_{j = 0}^{\color{#f00}{\infty}}\pars{-1}^{\,j}{q \choose j}\
\overbrace{{-p - q + i- 1 \choose i - j}\pars{-1}^{i - j}}^{\ds{p + q - j \choose i - j}}
\\[5mm] = &\
\pars{-1}^{i}\sum_{j = 0}^{\infty}{q \choose j}
\braces{\vphantom{\huge A}\bracks{z^{i - j}}\pars{1 + z}^{-p - q + i - 1}} =
\pars{-1}^{i}\bracks{z^{i}}\braces{\vphantom{\huge A}%
\pars{1 + z}^{-p - q + i - 1}\
\overbrace{\sum_{j = 0}^{\infty}{q \choose j}z^{j}}^{\ds{\pars{1 + z}^{q}}}}
\\[5mm] = &\
\pars{-1}^{i}\bracks{z^{i}}\pars{1 + z}^{-p + i - 1} =
\pars{-1}^{i}{-p + i - 1 \choose i}
\\[5mm] = &\
\pars{-1}^{i}\
\underbrace{{-\bracks{-p + i - 1} + i - 1 \choose i}\pars{-1}^{i}}
_{\ds{-p + i - 1 \choose i}}\ =\
\bbx{\ds{p \choose i}}
\end{align}
