= Solution
Put $r=k-1$. The linear-reproduction request requires $k\ge2$, which we use for that final step. Since $\psi_i=\omega_i/r!$, the useful <normalized Marsden dual functional> is
$$
\boxed{\lambda_i(p)=\sum_{j=0}^{r}(-1)^j p^{(j)}(x)\,
\psi_i^{(r-j)}(x)}.
$$
There is no further factor $1/r!$ outside this sum: it has already been included in $\psi_i$.
To derive the formula directly from <Marsden's identity>, <Taylor's theorem> for the <polynomial> $p$ around the arbitrary point $x$ gives
$$
p(t)=\sum_{j=0}^r\frac{(-1)^j}{j!}p^{(j)}(x)(x-t)^j.
$$
On the other hand,
$$
\partial_x^{\,r-j}(x-t)^r=\frac{r!}{j!}(x-t)^j.
$$
Differentiate <Marsden's identity> $r-j$ times in $x$, multiply by $(-1)^jp^{(j)}(x)/r!$, and sum over $j$. The left side becomes $p(t)$, while the right side becomes
$$
\boxed{p(t)=\sum_{i=1}^n\lambda_i(p)N_i(t)},\qquad t\in[t_k,t_{n+1}].
$$
Only finite sums and derivatives of <polynomials> are involved.
Independence of the auxiliary point does not require an assumption about uniqueness of an expansion. Differentiate the formula for $\lambda_i$ itself:
$$
\lambda_i'(x)=\sum_{j=0}^r(-1)^j
\left[p^{(j+1)}(x)\psi_i^{(r-j)}(x)
+p^{(j)}(x)\psi_i^{(r-j+1)}(x)\right]=0.
$$
The first summand at $j=r$ and the second at $j=0$ vanish because both <polynomials> have degree at most $r$. All remaining terms cancel after shifting the index by one. Therefore $\lambda_i(p)$ is a constant in $x$, and it is visibly a <linear functional> of $p$.
For $p(t)=a+bt$, only $j=0,1$ remain. The leading two coefficients of the monic knot <polynomial> give
$$
\psi_i^{(r)}(x)=1,\qquad
\psi_i^{(r-1)}(x)=x-\frac1r\sum_{\ell=1}^r t_{i+\ell}
=x-t_i^*.
$$
It follows that
$$
\lambda_i(p)=a+bx-b(x-t_i^*)=a+bt_i^*=p(t_i^*).
$$
Substituting into the expansion proves
$$
\boxed{p(t)=\sum_{i=1}^n p(t_i^*)N_i(t)}
$$
on the basic knot interval. Thus the <Greville abscissae> are exactly the sampling coefficients that reproduce linear <polynomials>; the constant case also gives the <subpartition of unity for B-splines> with equality on this interval.
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