= Solution
For fixed $t$, put $g_t(u)=(u-t)_+^{k-1}$, a <truncated power function>. Begin with distinct increasing knots. The two <interpolation polynomials> $\ell_i(\cdot,t)$ and $\ell_{i+1}(\cdot,t)$ agree at their $k-1$ shared knots. Their difference has degree at most $k-1$, so
$$
\ell_{i+1}(x,t)-\ell_i(x,t)=c_i(t)\prod_{j=1}^{k-1}(x-t_{i+j})
=c_i(t)\omega_i(x).
$$
For $k=1$ the empty product is one and the same assertion holds for constant interpolants. The <leading coefficient> of a degree-at-most-$k-1$ interpolant on $k$ nodes is its order-$k-1$ <divided difference>, by the <Newton interpolation polynomial>. Therefore
$$
c_i(t)=[t_{i+1},\ldots,t_{i+k}]g_t-[t_i,\ldots,t_{i+k-1}]g_t
=(t_{i+k}-t_i)[t_i,\ldots,t_{i+k}]g_t=N_i(t).
$$
This proves the <Lee interpolation identity>
$$
\boxed{\omega_i(x)N_i(t)=\ell_{i+1}(x,t)-\ell_i(x,t).}
$$
The identity holds for every real $x,t$, with the usual convention for the order-zero truncated power. Admissible repeated knots are handled by the corresponding confluent <divided differences> and consistent one-sided limits at breakpoints; taking these limits preserves the <Lee interpolation identity>.
Sum over $i$ to telescope:
$$
\sum_{i=1}^n\omega_i(x)N_i(t)=\ell_{n+1}(x,t)-\ell_1(x,t).
$$
If $t_k<t<t_{n+1}$, the first $k$ knot values of $g_t$ are zero, so $\ell_1=0$. At the last $k$ knots $t_{n+1},\ldots,t_{n+k}$, the function $g_t$ agrees with the <polynomial> $(u-t)^{k-1}$, so uniqueness of <polynomial interpolation> gives $\ell_{n+1}(x,t)=(x-t)^{k-1}$. Thus the <Marsden identity> is
$$
\boxed{(x-t)^{k-1}=\sum_{i=1}^n\omega_i(x)N_i(t),
\qquad t_k<t<t_{n+1},\quad x\in\mathbb R.}
$$
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