= Solution
Write the strictly increasing local <spline knots> as $u_j=t_{i+j}$, $0\le j\le k$. The explicit formula for a <divided difference> gives
$$
M_i(t)=k\sum_{j=0}^k
\frac{(u_j-t)_+^{k-1}}
{\prod_{\ell\ne j}(u_j-u_\ell)}.
$$
Each summand is a <truncated power function> of $t$: it is a <polynomial> on either side of its knot $u_j$, and for $k\ge2$ is globally $C^{k-2}$. Hence $M_i$ is a <piecewise polynomial function> of degree at most $k-1$, with these <spline knots> and at least this global smoothness.
For $t\ge u_k$, all summands vanish. For $t<u_0$, all knot values agree with those of the ordinary <polynomial> $(s-t)^{k-1}$ in the divided-difference variable $s$. Its order-$k$ <divided difference> is zero, because its degree is less than $k$. Thus $M_i$ also vanishes to the left of $u_0$.
The closed support is exactly the indicated interval, rather than merely contained in it. For $u_0<t<u_1$, subtracting the omitted $j=0$ term from the zero divided difference of $(s-t)^{k-1}$ gives
$$
M_i(t)=\frac{k(t-u_0)^{k-1}}{\prod_{\ell=1}^k(u_\ell-u_0)}>0.
$$
For $u_{k-1}<t<u_k$, only the last truncated-power term survives, giving
$$
M_i(t)=\frac{k(u_k-t)^{k-1}}{\prod_{\ell=0}^{k-1}(u_k-u_\ell)}>0.
$$
There are therefore nonzero values arbitrarily close to either endpoint, proving
$$
\boxed{\operatorname{supp}M_i=[t_i,t_{i+k}],\qquad M_i\in C^{k-2}}.
$$
At each simple knot, the $(k-1)$st derivative has a nonzero jump from exactly one truncated-power summand, so this is also the exact global smoothness. For order $k=1$, the function is instead the normalized interval indicator; there is no assertion of classical continuity, and the notation $C^{k-2}$ is only the customary formal spline smoothness notation. This distinction is part of <simple-knot B-spline regularity>.
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