A general formula to generate functions of power series I have been playing with Maclaurin series lately, I have been able to come across this:
$\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-x^5...$
$\dfrac{1}{(1+x)^2}=1-2x+3x^2-4x^3+5x^4-6x^5+7x^6...$
I found out by accident that:
$\dfrac{1-x}{(1+x)^3}=1-2^2x+3^2x^2-4^2x^3+5^2x^4+6^2x^5...$
I found in an old paper of Euler that this can continue on with these functions:
$\dfrac{1-4x+x^2}{(1+x)^4}=1-2^3x+3^3x^2-4^3x^3+5^3x^4+6^3x^5...$
$\dfrac{1-11x+11x^2-x^3}{(1+x)^5}=1-2^4x+3^4x^4-4^4x^3+5^4x^4+6^4x^5...$
$\dfrac{1-26x+66x^2-26x^3+x^4}{(1+x)^5}=1-2^5x+3^5x^4-4^5x^3+5^5x^4+6^5x^5...$
$\dfrac{1-57x+320x^2-302x^3+57x^4-x^5}{(1+x)^5}=1-2^6x+3^6x^4-4^6x^3+5^6x^4+6^6x^5...$
and so on...
Is there a general formula to generate the functions on the right hand side? How did Euler calculate these series? I have to say I deeply respect him since only he and Ramanujan know how to play with series.

The most efficient way to obtain such formulae is to compute the Newton series for $k$-th powers, and then use the fact that $1/(1-x)^{k+1} = \sum_{n=0}^∞ \binom{n+k}{k} x^n$ for $|x| < 1$, which is easy to prove by induction (or by observing that the coefficients for the series are a column of Pascal's triangle).
For example, we can easily get $n^3 = 1 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}$, as shown in the linked post, and hence
  $\sum_{n=0}^∞ (n+1)^3 x^n = \sum_{n=0}^∞ \left( 1 \binom{n+1}{1} + 6 \binom{n+1}{2} + 6 \binom{n+1}{3} \right) x^n$
    $ = \sum_{n=0}^∞ \left( 1 \binom{n+1}{1} + 6 \binom{n+2}{2} x + 6 \binom{n+3}{3} x^2 \right) x^n$   [since $\binom{n}{k} = 0$ for $0 ≤ n < k$]
    $ = 1/(1-x)^2 + 6x/(1-x)^3 + 6x^2/(1-x)^4$   [for $|x| < 1$]
    $ = \left( (1-x)^2 + 6x(1-x) + 6x^2 \right) / (1-x)^4$
    $ = \left( 1 + 4x + x^2 \right) / (1-x)^4$
(Substituting $x$ with $-x$ gives the first series cited in the question from "an old paper of Euler".)
