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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 $a_{i,j} = \binom{i+j}{i, j}$.

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    I think that in the first identity, you need $char(F)=0$ so that the binomial coefficients are defined (notice that you have division in the definition).2011-10-13
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    Of course, binomial coefficients can also be defined by Pascal's triangle, in which case it's just adding up 1's. Actually, in any case, it's within $\mathbb{Z}$, so you're good.2011-10-13
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    oh, yeah. sorry, I was a bit stupid for a moment...2011-10-13

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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 $k$, $0\le kk$ 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

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

I'm afraid this was just another(?) way of deriving that formula from Waterhouse's paper. Sorry, I cannot check that paper right now.

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    The first identity is, of course, _also_ the binomial formula in disguise (with negative exponents).2011-10-13
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    This was the way Waterhouse proved it - exploiting Fermat's Little Theorem and the equality of binomial coefficients you've observed (yet he was able to avoid Wilson's theorem elegantly). But what I'm looking for is why there are 2 identities describing the same quantity - $(1-x)^{-j}$ - and why they are different. As you've mentioned, it has a p-adic feeling to it, but I don't have enough p-adic education.2011-10-14
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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.