[proofplan] The proof is a direct specialization of Minkowski polynomiality for the volume of a Minkowski linear combination of convex bodies. We apply the two-body polynomiality formula to the pair $(K,B)$ with coefficients $1$ and $r$. The resulting degree-$n$ polynomial has one term for each ordered placement of copies of $K$ and $B$ inside the mixed volume; symmetry of mixed volume groups the terms with exactly $j$ copies of $B$ into $\binom{n}{j}$ equal contributions. The definition of $W_j(K)$ then gives precisely the displayed Steiner polynomial. [/proofplan]

[step:Apply Minkowski polynomiality to the pair $K$ and $B$]Since $K$ is a convex body and $B=\overline{B}(0,1)$ is also a convex body in $\mathbb{R}^n$, the two-body form of Minkowski polynomiality applies to the Minkowski linear combination $sK+tB$ for $s,t \geq 0$ (citing a result not yet in the wiki: Minkowski polynomiality for mixed volumes). For each $m \in \{1,\dots,n\}$, define the coefficient selector map $a_m: \{K,B\} \to [0,\infty)$ by $a_m(K)=s$ and $a_m(B)=t$. Thus there exist mixed-volume coefficients such that \begin{align*} \operatorname{Vol}_n(sK+tB) = \sum_{i_1,\dots,i_n \in \{K,B\}} V(i_1,\dots,i_n)\, a_1(i_1)\cdots a_n(i_n), \end{align*} where each $i_m$ is either $K$ or $B$. Set $s=1$ and $t=r$. Since $K+rB=1\cdot K+r\cdot B$, this gives \begin{align*} \operatorname{Vol}_n(K+rB) = \sum_{i_1,\dots,i_n \in \{K,B\}} V(i_1,\dots,i_n)\, r^{\#\{m: i_m=B\}}. \end{align*}[/step]

[guided]The relevant structural theorem is Minkowski polynomiality: the volume of a Minkowski linear combination of convex bodies is a homogeneous polynomial of degree $n$ in the scalar coefficients, and its coefficients are mixed volumes. We may apply it here because both inputs are convex bodies: $K$ is a convex body by hypothesis, and $B=\overline{B}(0,1)$ is compact, convex, and has nonempty interior. Applying the two-body form to $sK+tB$ gives a sum over all ordered $n$-tuples whose entries are either $K$ or $B$. If an ordered tuple is $(i_1,\dots,i_n)$, then its contribution is the mixed volume $V(i_1,\dots,i_n)$ multiplied by one scalar coefficient for each entry: a factor $s$ for every occurrence of $K$ and a factor $t$ for every occurrence of $B$. Now choose $s=1$ and $t=r$. The set $1\cdot K+r\cdot B$ is exactly $K+rB$, so the polynomiality formula becomes \begin{align*} \operatorname{Vol}_n(K+rB) = \sum_{i_1,\dots,i_n \in \{K,B\}} V(i_1,\dots,i_n)\, r^{\#\{m: i_m=B\}}. \end{align*} The exponent of $r$ records how many copies of $B$ occur in the ordered mixed-volume term.[/guided]

[step:Group the ordered mixed-volume terms by the number of copies of $B$]Fix $j \in \{0,\dots,n\}$. The ordered $n$-tuples $(i_1,\dots,i_n)$ with exactly $j$ entries equal to $B$ are obtained by choosing the $j$ positions occupied by $B$, and hence there are $\binom{n}{j}$ such tuples. By symmetry of mixed volume in its $n$ arguments, every such ordered tuple has the same mixed volume: \begin{align*} V(i_1,\dots,i_n) = V(\underbrace{K,\dots,K}_{n-j\text{ copies}},\underbrace{B,\dots,B}_{j\text{ copies}}) = W_j(K). \end{align*} Therefore the total contribution of all terms with exactly $j$ copies of $B$ is \begin{align*} \binom{n}{j}W_j(K)r^j. \end{align*}[/step]

[guided]We now collect like powers of $r$. A term contributes to the coefficient of $r^j$ exactly when the corresponding ordered tuple $(i_1,\dots,i_n)$ contains exactly $j$ copies of $B$ and therefore $n-j$ copies of $K$. How many such ordered tuples are there? We choose the $j$ positions among the $n$ available positions where $B$ appears. This gives exactly $\binom{n}{j}$ tuples. For each of these tuples, the mixed volume is the same. The reason is that mixed volume is symmetric in its arguments: permuting the arguments does not change its value. Hence any ordered tuple with $n-j$ copies of $K$ and $j$ copies of $B$ has value \begin{align*} V(\underbrace{K,\dots,K}_{n-j\text{ copies}},\underbrace{B,\dots,B}_{j\text{ copies}}). \end{align*} By the definition of the quermassintegral used in the statement, this common value is $W_j(K)$. Thus all terms with exactly $j$ copies of $B$ combine into \begin{align*} \binom{n}{j}W_j(K)r^j. \end{align*}[/guided]

[step:Sum the grouped contributions to obtain the Steiner polynomial] Summing the grouped contributions over all possible values $j=0,\dots,n$ gives \begin{align*} \operatorname{Vol}_n(K+rB) = \sum_{j=0}^{n}\binom{n}{j}W_j(K)r^j. \end{align*} This is the asserted Steiner formula. The argument used only Minkowski polynomiality for convex bodies and symmetry of mixed volume, so no smoothness or regularity assumption on $\partial K$ is required. [/step]