= Solution
Use the usual strictly increasing <spline knot sequence>, and initially assume $k\ge2$. The explicit formula for the order-$k$ <divided difference> gives
$$
M_i(t)=k\sum_{j=i}^{i+k}\frac{(t_j-t)_+^{k-1}}{\prod_{\substack{\ell=i\\\ell\ne j}}^{i+k}(t_j-t_\ell)}.
$$
Each summand is a <truncated power function> in $t$, with its junction at $t_j$. It is a <piecewise polynomial function> of degree at most $k-1$ and has $k-2$ <continuous> <derivatives>. Finite summation preserves these properties.
For $t\ge t_{i+k}$ all summands vanish. For $t\le t_i$, all <spline knot> data come from the <polynomial> $u\mapsto(u-t)^{k-1}$. An order-$k$ <divided difference> annihilates a <polynomial> of smaller degree, so $M_i(t)=0$ there too. Thus the <support> is contained in $[t_i,t_{i+k}]$.
On the last open <spline knot> interval only the $j=i+k$ term survives; it is a nonzero multiple of $(t_{i+k}-t)^{k-1}$. On the first open <spline knot> interval, compare the truncated <spline knot> data with the untruncated <polynomial> data: only the $j=i$ value differs. Since the untruncated <divided difference> is zero, the result is a nonzero multiple of $(t_i-t)^{k-1}$. Consequently both outer pieces are nonzero, giving exact closed <support>. The <Cox-de Boor recurrence> used below also proves positivity throughout the open <support>. Therefore
$$
\boxed{M_i\in C^{k-2}(\mathbb R),\qquad\operatorname{supp}M_i=[t_i,t_{i+k}],\qquad\deg\text{ each piece}\le k-1.}
$$
This is <simple-knot B-spline regularity>. For $k=1$ the <B-splines> are interval indicators; they are piecewise constant and need not be <continuous>. The notation $C^{k-2}$ should then be read as no <continuity> requirement, not as a classical negative-order differentiability space.
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