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)?

The first one is just inclusion-exclusion in the following way:
Take the set $[p+q]=\{1,\cdots ,p+q\},$ so you want to take $i$ elements from those such that they all belong to $[p].$ By definition you just restrict yourself to the set $[p]$ and hence there are $\binom{p}{i}$, but on the other hand it is the same as this $$|T\setminus \bigcup _{j=1}^q A_{p+j}|,$$
where $T$ is take all possible subsets of size $i$ from $[p+q]$ which can be done in $\binom{p+q}{i}$ and $A_r = \{S\subset [p+q]:|S|=i \wedge r\in S\}$ (so we are taking out all sets that contain elements on $[p+q]\setminus [p]$.)
By inclusion-exclusion then $$|T\setminus \bigcup _{j=1}^q A_{p+j}|=|T|-\sum _{j=1}^q(-1)^{j-1}\sum _{X\in \binom{[p+q]\setminus [p]}{j}}|\bigcap _{y\in X} A_{y}|,$$
as seeing before, $|T|=\binom{p+q}{i},$ and $|A_r|=\binom{p+q-1}{i-1}$ and if you take $r_1,r_2\in [p+q]\setminus [p],$ $|A_{r_1}\cap A_{r_2}|=\binom{p+q-2}{i-2}$ because you have already chosen $2$, hence the intersections are homogeneous and then $$|T|-\sum _{j=1}^q(-1)^{j-1}\sum _{X\in \binom{[p+q]\setminus [p]}{j}}|\bigcap _{y\in X} A_{y}|=\binom{p+q}{i}-\sum _{j=1}^q(-1)^{j-1}\sum _{X\in \binom{[p+q]\setminus [p]}{j}}\binom{p+q-j}{i-j}=\binom{p+q}{i}-\sum _{j=1}^q(-1)^{j-1}\binom{q}{j}\binom{p+q-j}{i-j},$$
what is you identity.
The second one seems more challenging(but it suggests a probabilistic approach.)
Added: The second is a particular case of the first one. 
Notice that $\frac{\binom{i}{j}}{\binom{q-j}{i-j}}=\frac{(q-i)!i!}{j!(q-j)!}=\frac{\binom{q}{j}}{\binom{q}{i}},$ so on the LHS you get $$\binom{q-1}{i},$$
and then take $p=q-1$ in your first identity and the result follows.
