= Solution
For order $k\geq2$, set $g(u)=(u-t)_+^{k-2}$. Then $(u-t)g(u)=(u-t)_+^{k-1}$. Apply the given <divided difference> product identity with nodes $t_i,\ldots,t_{i+k}$ and multiply by the length $t_{i+k}-t_i$ of their knot interval:
$$
\begin{aligned}
N_{i,k}(t)
&=(t-t_i)[t_i,\ldots,t_{i+k-1}]g\\
&\quad +(t_{i+k}-t)[t_{i+1},\ldots,t_{i+k}]g.
\end{aligned}
$$
Using the definition of each lower-order normalized <B-spline>, the two <divided differences> are $N_{i,k-1}(t)/(t_{i+k-1}-t_i)$ and $N_{i+1,k-1}(t)/(t_{i+k}-t_{i+1})$. Hence the <Cox-de Boor recurrence> is
$$
\boxed{
N_{i,k}(t)=
\frac{t-t_i}{t_{i+k-1}-t_i}N_{i,k-1}(t)
+\frac{t_{i+k}-t}{t_{i+k}-t_{i+1}}N_{i+1,k-1}(t).}
$$
For completeness, the product identity itself follows from
$$
[t_0,\ldots,t_k](u-t)g(u)
=[t_0,\ldots,t_{k-1}]g+(t_k-t)[t_0,\ldots,t_k]g
$$
and the ordinary <divided difference> recurrence. Substitution yields the stated convex combination of the two order-$k-1$ <divided differences>.
The initial order-one <splines> are the indicators $N_{i,1}(t)=\mathbf1_{[t_i,t_{i+1})}(t)$. For repeated knots, a term with zero knot-span denominator is defined as zero; this agrees with the standard limiting basis convention. The recurrence uses the <partition of unity> normalization with $0\leq N_{i,k}\leq1$, not a separate rescaling of each <spline> to make its actual supremum exactly one. The original PDF recurrence has the denominators above; several of these indices and signs were corrupted in the converted TeX.
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