Calculate Power Series Coefficients Given a polynomial \$p(x)\$ with integral coefficients and a constant term of \$p(0) = \pm 1\$, and a non-negative integer \$N\$, return the \$N\$-th coefficient of the power series (sometimes called &quot;Taylor series&quot;) of \$f(x) = \frac{1}{p(x)}\$ developed at \$x_0 = 0\$, i.e., the coefficient of the monomial of degree \$N\$.
The given conditions ensure that the power series exist and that the its coefficients are integers.
Details
As always the polynomial can be accepted in any convenient format, e.g. a list of coefficients, for instance \$p(x) = x^3-2x+5\$ could be represented as [1,0,-2,5].
The power series of a function \$f(x)\$ developed at \$0\$ is given by
$$f(x) = \sum_{k=0}^\infty{\frac{f^{(n)}(0)}{n!}x^n}$$
and the \$N\$-th coefficient (the coefficient of \$x^N\$) is given by
$$\frac{f^{(N)}}{N!}$$
where \$f^{(n)}\$ denotes the \$n\$-th derivative of \$f\$
Examples

The polynomial \$p(x) = 1-x\$ results in the geometric series \$f(x) = 1 + x + x^2 + ...\$ so the output should be \$1\$ for all \$N\$.

\$p(x) = (1-x)^2 = x^2 - 2x + 1\$ results in the derivative of the geometric series \$f(x) = 1 + 2x + 3x^2 + 4x^3 + ...\$, so the output for \$N\$ is \$N+1\$.

\$p(x) = 1 - x - x^2\$ results in the generating function of the Fibonacci sequence \$f(x) = 1 + x + 2x^2 + 3x^3 + 5x^4 + 8x^5 + 13x^6 + ...\$

\$p(x) = 1 - x^2\$ results in the generating function of \$1,0,1,0,...\$ i.e. \$f(x) = 1 + x^2 + x^4 + x^6 + ...\$

\$p(x) = (1 - x)^3 = 1 -3x + 3x^2 - x^3\$ results in the generating function of the triangular numbers \$f(x) = 1 + 3x + 6x^6 + 10x^3 + 15x^4 + 21x^5 + ...\$ that means the \$N\$-th coefficient is the binomial coefficient \$\binom{N+2}{N}\$

\$p(x) = (x - 3)^2 + (x - 2)^3 = 1 + 6x - 5x^2 + x^3\$ results in \$f(x) = 1 - 6x + 41x^2 - 277x^3 + 1873x4 - 12664x^5 + 85626x^6 - 57849x^7 + \dots\$
Mathematica, 24 23 bytes
Saved 1 byte thanks to Greg Martin
D[1/#2,{x,#}]/#!/.x-&gt;0&amp;

Pure function with two arguments # and #2. Assumes the polynomial #2 satisfies PolynomialQ[#2,x]. Of course there's a built-in for this:
SeriesCoefficient[1/#2,{x,0,#}]&amp;