How to interpret this result modulo $(y-1)^{n+1}$?

Question

I recently discovered that the following identity is true:

$$ \boxed{\frac{\partial^{n+1}}{\partial x^{n+1}}\left(\frac{(xy-1)^n}{n!} \log \frac{1}{1-x}\right) \equiv \frac{y^{n+1}}{1-xy} \pmod{(y-1)^{n+1}}} $$

To add a bit more context, for any polynomial $f(x)$ of degree at most $n$, we could apply the linear functional $T(y^i) = f(i)$ to the formula above, and it will yield the result about the values of $f(x)$, specifically

$$ \frac{\partial^{n+1}}{\partial x^{n+1}}\left( \sum\limits_{i=0}^n \frac{(-1)^{n-i} f(i)}{i!(n-i)!} x^i \right) \left(\sum\limits_{j=1}^{\infty} \frac{x^j}{j} \right) = \sum\limits_{k=1}^\infty f(n+k) x^{k-1}. $$

It then allows, for a given $f(0), \dots, f(n)$ and integer $s$ to find $f(s),\dots, f(s+n)$ in a single convolution.

The result above can as well be obtained by somewhat mindless formula bash starting from the Lagrange interpolation formula, but I'd really like to know if there is any deeper meaning to the expression above in terms of $x$ and $y$.

Answer 1

$$\frac{\partial}{\partial x} \frac{(xy-1)^a}{(1-x)^b} = ay \frac{(xy-1)^{a-1}}{(1-x)^b} + b \frac{(xy-1)^a}{(1-x)^{b+1}}$$ from which it's easy to see either by induction or combinatorially that $$\frac{\partial^n}{\partial x^n} \frac{(xy-1)^a}{(1-x)^b} = \sum_{j=0}^n \binom{n}{j} a^{\underline{j}} b^{\overline{n-j}} y^j \frac{(xy-1)^{a-j}}{(1-x)^{b+n-j}}$$ and in particular when $a=n, b=1$ we have $$\frac{\partial^n}{\partial x^n} \frac{(xy-1)^n}{1-x} = \sum_{j=0}^n \binom{n}{j} \color{red} {n^{\underline{j}} 1^{\overline{n-j}}} y^j \frac{(xy-1)^{n-j}}{(1-x)^{\color{blue}1+n-j}} = \frac{\color{red}{n!}}{\color{blue}{1-x}} \sum_{j=0}^n \binom{n}{j} y^j \left(\frac{xy-1}{1-x}\right)^{n-j} = \frac{n!}{1-x} \left(y + \frac{xy-1}{1-x} \right)^n = \frac{n!}{1-x} \left(\frac{y-1}{1-x} \right)^n $$

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Then $$\begin{eqnarray*} \frac{\partial^{n+1}}{\partial x^{n+1}} \left( \frac{(xy-1)^n}{n!} \log \frac{1}{1-x}\right) &=& y \frac{\partial^n}{\partial x^n} \left(\frac{(xy-1)^{n-1} }{(n-1)!} \log \frac{1}{1-x}\right) + \frac{\partial^n}{\partial x^n} \frac{(xy-1)^n}{n!(1-x)} \\ &=& y \frac{\partial^n}{\partial x^n} \left(\frac{(xy-1)^{n-1} }{(n-1)!} \log \frac{1}{1-x}\right) + \frac{1}{1-x} \left(\frac{y-1}{1-x} \right)^n \\ &\vdots& \\ %&=& y^n \frac{\partial}{\partial x} \log \frac{1}{1-x} + y^{n-1} \frac{1}{1-x} \left(\frac{y-1}{1-x} \right) + \cdots + \frac{1}{1-x} \left(\frac{y-1}{1-x} \right)^n \\ &=& y^n \frac{1}{1-x} + y^{n-1} \frac{1}{1-x} \left(\frac{y-1}{1-x} \right) + \cdots + \frac{1}{1-x} \left(\frac{y-1}{1-x} \right)^n \\ %&=& \frac{y^n}{1-x} \left(1 + \left(\frac{y-1}{y-xy} \right) + \cdots + \left(\frac{y-1}{y-xy} \right)^n \right) \\ %&=& \frac{y^n}{1-x} \frac{1-\left(\frac{y-1}{y-xy} \right)^{n+1}}{1-\left(\frac{y-1}{y-xy} \right)} \\ %&=& \frac{y^{n+1}}{1-xy} \left(1-\left(\frac{y-1}{y-xy} \right)^{n+1}\right) \\ %&=& \frac{1}{1-xy} \left(y^{n+1} - \left(\frac{y-1}{1-x} \right)^{n+1}\right) \\ \end{eqnarray*}$$ Thus far I think it can be argued to have a combinatorial interpretation in terms of the number of times you differentiate the polynomial part before differentiating the log part, and then the number of times you differentiate the numerator and denominator of the resulting rational function; the final manipulation of the geometric series to obtain $$= \frac{1}{1-xy} \left(y^{n+1} - \left(\frac{y-1}{1-x} \right)^{n+1}\right)$$ is more purely algebraic.

I'm not sure to what extent you will find this satisfying, though. It seems to rely on some "coincidences", including $n^{\underline{j}} 1^{\overline{n-j}} = n!$ and $y + \frac{xy-1}{1-x} = \frac{y-1}{1-x}$.