 Solution

Differentiate the finite binomial theorem and set $z=1$. For $n\ge 1$ this givesFor $n=0$ the sum is $0$ directly.

For the alternating binomial moments, it is useful to differentiate in a form that introduces falling factorials. Write $(m{)}_j=m(m-1)\cdots (m-j+1)$, with $(m{)}_0=1$. Differentiating the binomial theorem $j$ times and multiplying by $z^j$ yields, for $0\le j\le n$,At $z=-1$, the right side vanishes when $j<n$, and equals $(-1{)}^{n}n!$ when $j=n$.

The polynomials $(X{)}_0,\dots ,(X{)}_r$ have degrees $0,\dots ,r$ and leading coefficient $1$, so successive cancellation of leading terms expresses $X^r$ as a linear combination of them. If $r<n$, every falling-factorial term has index $j<n$ and hence its alternating sum vanishes. If $r=n$, the coefficient of $(X{)}_n$ is $1$, and all lower-index terms still vanish. Therefore the alternating moments areFor $n=0$, the latter identity uses the constant polynomial $X^0=1$, so its value at $X=0$ is $1$; thus $S_0(0)=1$ as well. This makes the edge-case convention explicit rather than treating $0^0$ as an unrelated numerical expression.