= Solution
Differentiate the finite <binomial theorem> and set $z=1$. For $n\geq1$ this gives
$$
\boxed{\sum_{m=0}^n m\binom nm=n2^{n-1}}.
$$
For $n=0$ 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 $0\leq j\leq n$,
$$
\sum_{m=0}^n(m)_j\binom nm z^m=(n)_jz^j(1+z)^{n-j}.
$$
At $z=-1$, the right side vanishes when $j<n$, and equals $(-1)^nn!$ when $j=n$.
The <polynomials> $(X)_0,\ldots,(X)_r$ have degrees $0,\ldots,r$ 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 \b[the alternating moments are]
$$
\boxed{S_r(n)=0\quad(0\leq r<n),\qquad S_n(n)=(-1)^nn!}.
$$
For $n=0$, 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.
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