= Solution
Write $g_t(y)=(y-t)_+^{k-1}$. Begin with distinct increasing knots so ordinary <polynomial interpolation> is unambiguous. The two interpolants agree at the $k-1$ common knots $t_{i+1},\ldots,t_{i+k-1}$. Their difference is consequently a scalar multiple of the monic <polynomial> $\omega_i$ of degree $k-1$.
The leading coefficient of an interpolant is its highest <divided difference>. Thus this scalar is
$$
[t_{i+1},\ldots,t_{i+k}]g_t-[t_i,\ldots,t_{i+k-1}]g_t
=(t_{i+k}-t_i)[t_i,\ldots,t_{i+k}]g_t=N_i(t).
$$
The equality is precisely the divided-difference recurrence. This proves \b[the <Lee interpolation identity>]:
$$
\boxed{\ell_{i+1}(x,t)-\ell_i(x,t)=\omega_i(x)N_i(t).}
$$
For $k=1$ the common-knot product is empty; the same argument is a difference of constants. Interpret $(y-t)_+^0$ as $1_{y>t}$ to obtain the usual half-open order-one <B-splines>. Repeated knots use the standard confluent or limiting interpretation, where the interpolation data are defined.
Sum over $i=1,\ldots,n$. The right side telescopes to $\ell_{n+1}(x,t)-\ell_1(x,t)$. When $t_k<t<t_{n+1}$, the first interpolation nodes all lie below $t$, so $\ell_1=0$. The last interpolation nodes all lie above $t$, so $\ell_{n+1}$ interpolates the <polynomial> $(x-t)^{k-1}$ of degree $k-1$ and equals it identically. Hence \b[the <Marsden identity> follows]:
$$
\boxed{(x-t)^{k-1}=\sum_{i=1}^n\omega_i(x)N_i(t),\qquad t_k<t<t_{n+1}.}
$$
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