[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*}
[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]
[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*}
[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]
[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]
