= Solution
For any admissible $f$, part (a) rewrites the constraints as
$$
(M_i,f^{(k)})=k!\gamma_i=(M_i,s),
\qquad i=1,\ldots,n.
$$
Hence $f^{(k)}-s$ is orthogonal to every $M_i$ and therefore to their span, which contains $s$. The <Pythagorean theorem in an inner-product space> gives
$$
\|f^{(k)}\|_2^2
=\|s\|_2^2+\|f^{(k)}-s\|_2^2
\geq\|s\|_2^2.
$$
Choose any $k$-fold antiderivative $\sigma$ of $s$. The identity from part (a) and $(M_i,s)=k!\gamma_i$ show that $\sigma$ satisfies all prescribed divided differences, and equality holds in the norm bound. Therefore
$$
\boxed{\sigma^{(k)}=s}
$$
characterizes the minimizers. They are unique up to addition of an arbitrary polynomial in $\mathcal P_{k-1}$, which changes neither the order-$k$ divided differences nor the $k$th derivative.
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