Recently, I found two $q$-binomial identities
$$ \sum_{p=0}^b \binom{a}{p}_q q^{\binom{p}{2}} (-1)^p = \binom{b-a}{b}_q q^{ab}, $$ and
$$ \sum_{p=a}^b \binom{p-1}{a-1}_q q^{-ap} = \binom{b}{a}_q q^{-ab}. $$
Even though they look very simple, and their special case when $q=1$ are well-known (see this and that), I couldn't find them anywhere.
Question: I'm looking for references containing both of these identities (or equivalents).
After further research, I've found the recent reference
M. Shattuck. On some relations satisfied by the $p, q$-binomial coefficient. Šiauliai Math. Semin., 6(14):69–84, 2011.
which contains generalization of both identities. They are respectively (and at a renaming of the variables away) Identity 3.9 with $q=1$ and $m=i=0$ and Identity 3.1 with $p=1$.
Here is a proof which you may be interested in seeing. Recall the Cauchy Binomial Theorem which says that
$${n\choose k}_q = q^{-k(k-1)/2} [t^k] \prod_{r=0}^{n-1} (1+q^rt).$$
With the first identity we seek to verify where $a\ge b$
$$\sum_{p=0}^b {a\choose p}_q q^{p(p-1)/2} (-1)^p = {b-a\choose b}_q q^{ab}.$$
We obtain for the sum
$$\sum_{p=0}^b (-1)^p [t^p] \prod_{r=0}^{a-1} (1+q^rt) = [t^b] \frac{1}{1-t} \prod_{r=0}^{a-1} (1-q^rt) = [t^b] \prod_{r=1}^{a-1} (1-q^rt).$$
Now from the definition we have
$${-m\choose r}_q = \frac{(1-q^{-m})\cdots(1-q^{-m-r+1})}{(1-q)\cdots(1-q^r)} \\ = q^{-\sum_{p=m}^{m+r-1} p} \frac{(q^m-1)\cdots(q^{m+r-1}-1)}{(1-q)\cdots(1-q^r)} \\ = (-1)^r q^{-\sum_{p=m}^{m+r-1} p} {m+r-1\choose r}_q.$$
This is the familiar upper negation. We get for the closed form
$$(-1)^b q^{-\sum_{p=a-b}^{a-1} p} q^{ab} {a-1\choose b}_q \\ = (-1)^b q^{-\sum_{p=a-b}^{a-1} p} q^{ab} q^{-b(b-1)/2} [t^b] \prod_{r=0}^{a-2} (1+q^rt) \\ = q^b [t^b] \prod_{r=0}^{a-2} (1-q^rt) = [t^b] \prod_{r=0}^{a-2} (1-q^{r+1}t).$$
This is the claim.
Here we seek with $b\ge a$ that
$$\sum_{p=a}^b {p-1\choose a-1}_q q^{-ap} = {b\choose a}_q q^{-ab}.$$
We cite a generalization of the Cauchy Binomial Theorem which is
$${n+k-1\choose k}_q = [t^k] \prod_{r=0}^{n-1} \frac{1}{1-q^rt}.$$
Working with the coefficient in the sum we write
$${p-1\choose p-a}_q = {p-a+a-1\choose p-a}_q = [t^{p-a}] \prod_{r=0}^{a-1} \frac{1}{1-q^rt}.$$
We get for the sum
$$\sum_{p=a}^b q^{-ap} [t^{p-a}] \prod_{r=0}^{a-1} \frac{1}{1-q^rt} = q^{-a^2} \sum_{p=0}^{b-a} q^{-ap} [t^p] \prod_{r=0}^{a-1} \frac{1}{1-q^rt} \\ = q^{-a^2} \sum_{p=0}^{b-a} [t^p] \prod_{r=0}^{a-1} \frac{1}{1-q^{r-a}t} = q^{-a^2} [t^{b-a}] \frac{1}{1-t} \prod_{r=0}^{a-1} \frac{1}{1-q^{r-a}t} \\ = q^{-a^2} [t^{b-a}] \prod_{r=0}^a \frac{1}{1-q^{r-a}t} = q^{-a^2} q^{-a(b-a)} [t^{b-a}] \prod_{r=0}^a \frac{1}{1-q^{rt}} \\ = q^{-ab} {a+b-a\choose b-a}_q = q^{-ab} {b\choose a}_q.$$
Once more we have the claim.
