A resemblance between 2 binomial identities - why? Let $F$ be any field (or even ring). The following formal power series identity (i.e., equality in $F[[x]]$) holds for any $j \ge 0$:
$$(1-x)^{-j} = \sum_{i \ge 0} \binom{i +j -1}{i} x^i $$
The other identity is the following: let $F_{p}$ be a finite field with $p$ elements. The following holds for each $1 \neq x \in F_{p}$:
$$(1-x)^{-j} = \sum_{i \ge 0}^{p-1} \binom{i +j }{i} x^i$$
Are the identities equivalent or related in some way? I feel that the first identity is more formal and concerns equality of coefficients (after multiplying both sides by $(1-x)^j$), yet the second identity is not only formal: $x$ can be plugged in and both sides give an element in the field.
Any insights are welcome - I'm not sure what I'm expecting.
Context: I've encountered the first identity - which is quite useful - while studying generating functions. I've learned of the second identity in a paper by Waterhouse about a matrix with elements $a_{i,j} = \binom{i+j}{i, j}$.
 A: A good question. As you observed the first series is strictly formal over $F_p$, and we cannot really substitute anything in place of $x$ there. A $p$-adic interpretation may be possible, but I don't see it now.
The way I look at it, the second identity is the binomial formula in disguise. Let me elaborate on that. In the field $F_p$ we have the familiar Wilson's theorem:
$$ (p-1)!=-1.$$
Let us fix and integer $k$, $0\le k<p$. By writing the factors $\ell >k$ in $(p-1)!$ in the form $\ell=-(p-\ell)$ we can rewrite Wilson's theorem as stating that
$$
k!(p-1-k)!(-1)^{p-1-k}=-1 \Leftrightarrow \frac{1}{(p-1-k)!}=(-1)^{p-k}k!.
$$
If $x\neq1$, then $(1-x)$ is a non-zero element of the field $F_p$, and we can write for any $j$ in the range $0\le j<p$
$$
(1-x)^{-j}=(1-x)^{p-1-j}=\sum_{i=0}^{p-1-j}(-1)^i{p-1-j\choose i}x^i.
$$
Using the above consequence of Wilson's theorem we can rewrite this binomial coefficient as
$$
{p-1-j\choose i}=\frac{(p-1-j)!}{i!(p-1-i-j)!}=\frac{(-1)^{i+j}(i+j)!}{i! (-1)^j j!}=(-1)^i{i+j\choose i}.
$$
Putting these two bits together we get
$$
(1-x)^{-j}=\sum_{i=0}^{p-1-j}{i+j\choose i}x^i.
$$
It is easy to see that when $i$ is in the range $p-j\le i <p$, the binomial coefficient
${i+j\choose i}$ is divisible by $p$. The identity that you listed follows from this.
I'm afraid this was just another(?) way of deriving that formula from Waterhouse's paper. Sorry, I cannot check that paper right now.
A: The link is that you take the first identity and lump together all exponents of $x$ that differ by multiples $p-1$. The shift by 1 corresponds to the sum formula for binomial coefficients.
But checking the details is not easier than the other proof.
